message stringlengths 2 22.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 16 109k | cluster float64 1 1 | __index_level_0__ int64 32 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
To get on the Shinkansen, you need two tickets, a "ticket" and a "limited express ticket". These are separate tickets because some of the routes may not use the Shinkansen, but for routes that use only the Shinkansen, one ticket can be used as both a ticket and a limited express ticket. It may also be issued.
Automatic ticket gates must read these tickets and open the gate only when the correct ticket is inserted. Create a program that determines whether one "ticket" and one "limited express ticket" or both, or one "boarding / limited express ticket" has been inserted, and determines whether the door of the automatic ticket gate is open or closed. please do it.
input
The input is given in the following format.
b1 b2 b3
The input consists of one line and contains three integers separated by one space. b1 indicates the state in which the "ticket" is inserted, b2 indicates the state in which the "limited express ticket" is inserted, and b3 indicates the state in which the "boarding / limited express ticket" is inserted. The turned-in state is represented by 0 or 1, where 0 indicates a non-charged state, and 1 indicates a loaded state. However, the combination of expected input states is limited to the following cases.
Input | Input state | Door operation for input
--- | --- | ---
1 0 0 | Insert only "ticket" | Close
0 1 0 | Only "Limited Express Ticket" is introduced | Close
1 1 0 | Introducing "tickets" and "express tickets" | Open
0 0 1 | Introducing "Boarding / Limited Express Tickets" | Open
0 0 0 | No input | Close
output
Outputs Open or Close, which indicates the opening and closing of the automatic ticket gate, on one line.
Example
Input
0 0 1
Output
Open | instruction | 0 | 52,957 | 1 | 105,914 |
"Correct Solution:
```
b1,b2,b3=map(int,input().split())
if (b1==1 and b2==1 and b3==0) or (b1==0 and b2==0 and b3==1):
print("Open")
else:
print("Close")
``` | output | 1 | 52,957 | 1 | 105,915 |
Provide a correct Python 3 solution for this coding contest problem.
To get on the Shinkansen, you need two tickets, a "ticket" and a "limited express ticket". These are separate tickets because some of the routes may not use the Shinkansen, but for routes that use only the Shinkansen, one ticket can be used as both a ticket and a limited express ticket. It may also be issued.
Automatic ticket gates must read these tickets and open the gate only when the correct ticket is inserted. Create a program that determines whether one "ticket" and one "limited express ticket" or both, or one "boarding / limited express ticket" has been inserted, and determines whether the door of the automatic ticket gate is open or closed. please do it.
input
The input is given in the following format.
b1 b2 b3
The input consists of one line and contains three integers separated by one space. b1 indicates the state in which the "ticket" is inserted, b2 indicates the state in which the "limited express ticket" is inserted, and b3 indicates the state in which the "boarding / limited express ticket" is inserted. The turned-in state is represented by 0 or 1, where 0 indicates a non-charged state, and 1 indicates a loaded state. However, the combination of expected input states is limited to the following cases.
Input | Input state | Door operation for input
--- | --- | ---
1 0 0 | Insert only "ticket" | Close
0 1 0 | Only "Limited Express Ticket" is introduced | Close
1 1 0 | Introducing "tickets" and "express tickets" | Open
0 0 1 | Introducing "Boarding / Limited Express Tickets" | Open
0 0 0 | No input | Close
output
Outputs Open or Close, which indicates the opening and closing of the automatic ticket gate, on one line.
Example
Input
0 0 1
Output
Open | instruction | 0 | 52,958 | 1 | 105,916 |
"Correct Solution:
```
b1, b2, b3 = map(int,input().split())
if b1 == 1 and b2 == 1:
print('Open')
elif b3 == 1:
print('Open')
else:
print('Close')
``` | output | 1 | 52,958 | 1 | 105,917 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To get on the Shinkansen, you need two tickets, a "ticket" and a "limited express ticket". These are separate tickets because some of the routes may not use the Shinkansen, but for routes that use only the Shinkansen, one ticket can be used as both a ticket and a limited express ticket. It may also be issued.
Automatic ticket gates must read these tickets and open the gate only when the correct ticket is inserted. Create a program that determines whether one "ticket" and one "limited express ticket" or both, or one "boarding / limited express ticket" has been inserted, and determines whether the door of the automatic ticket gate is open or closed. please do it.
input
The input is given in the following format.
b1 b2 b3
The input consists of one line and contains three integers separated by one space. b1 indicates the state in which the "ticket" is inserted, b2 indicates the state in which the "limited express ticket" is inserted, and b3 indicates the state in which the "boarding / limited express ticket" is inserted. The turned-in state is represented by 0 or 1, where 0 indicates a non-charged state, and 1 indicates a loaded state. However, the combination of expected input states is limited to the following cases.
Input | Input state | Door operation for input
--- | --- | ---
1 0 0 | Insert only "ticket" | Close
0 1 0 | Only "Limited Express Ticket" is introduced | Close
1 1 0 | Introducing "tickets" and "express tickets" | Open
0 0 1 | Introducing "Boarding / Limited Express Tickets" | Open
0 0 0 | No input | Close
output
Outputs Open or Close, which indicates the opening and closing of the automatic ticket gate, on one line.
Example
Input
0 0 1
Output
Open
Submitted Solution:
```
b1,b2,b3 = map(int,input().split())
if b3 == 1 :
print("Open")
elif b1 == b2 == 1 :
print("Open")
else :
print("Close")
``` | instruction | 0 | 52,959 | 1 | 105,918 |
Yes | output | 1 | 52,959 | 1 | 105,919 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To get on the Shinkansen, you need two tickets, a "ticket" and a "limited express ticket". These are separate tickets because some of the routes may not use the Shinkansen, but for routes that use only the Shinkansen, one ticket can be used as both a ticket and a limited express ticket. It may also be issued.
Automatic ticket gates must read these tickets and open the gate only when the correct ticket is inserted. Create a program that determines whether one "ticket" and one "limited express ticket" or both, or one "boarding / limited express ticket" has been inserted, and determines whether the door of the automatic ticket gate is open or closed. please do it.
input
The input is given in the following format.
b1 b2 b3
The input consists of one line and contains three integers separated by one space. b1 indicates the state in which the "ticket" is inserted, b2 indicates the state in which the "limited express ticket" is inserted, and b3 indicates the state in which the "boarding / limited express ticket" is inserted. The turned-in state is represented by 0 or 1, where 0 indicates a non-charged state, and 1 indicates a loaded state. However, the combination of expected input states is limited to the following cases.
Input | Input state | Door operation for input
--- | --- | ---
1 0 0 | Insert only "ticket" | Close
0 1 0 | Only "Limited Express Ticket" is introduced | Close
1 1 0 | Introducing "tickets" and "express tickets" | Open
0 0 1 | Introducing "Boarding / Limited Express Tickets" | Open
0 0 0 | No input | Close
output
Outputs Open or Close, which indicates the opening and closing of the automatic ticket gate, on one line.
Example
Input
0 0 1
Output
Open
Submitted Solution:
```
a, b, c = map(int, input().split())
if a + b == 2 or c == 1:
print('Open')
else:
print('Close')
``` | instruction | 0 | 52,960 | 1 | 105,920 |
Yes | output | 1 | 52,960 | 1 | 105,921 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To get on the Shinkansen, you need two tickets, a "ticket" and a "limited express ticket". These are separate tickets because some of the routes may not use the Shinkansen, but for routes that use only the Shinkansen, one ticket can be used as both a ticket and a limited express ticket. It may also be issued.
Automatic ticket gates must read these tickets and open the gate only when the correct ticket is inserted. Create a program that determines whether one "ticket" and one "limited express ticket" or both, or one "boarding / limited express ticket" has been inserted, and determines whether the door of the automatic ticket gate is open or closed. please do it.
input
The input is given in the following format.
b1 b2 b3
The input consists of one line and contains three integers separated by one space. b1 indicates the state in which the "ticket" is inserted, b2 indicates the state in which the "limited express ticket" is inserted, and b3 indicates the state in which the "boarding / limited express ticket" is inserted. The turned-in state is represented by 0 or 1, where 0 indicates a non-charged state, and 1 indicates a loaded state. However, the combination of expected input states is limited to the following cases.
Input | Input state | Door operation for input
--- | --- | ---
1 0 0 | Insert only "ticket" | Close
0 1 0 | Only "Limited Express Ticket" is introduced | Close
1 1 0 | Introducing "tickets" and "express tickets" | Open
0 0 1 | Introducing "Boarding / Limited Express Tickets" | Open
0 0 0 | No input | Close
output
Outputs Open or Close, which indicates the opening and closing of the automatic ticket gate, on one line.
Example
Input
0 0 1
Output
Open
Submitted Solution:
```
a, b, c= map(int,input().split())
if a==1. and b==0. and c==0:
print("Close")
elif a==0. and b==1. and c==0:
print("Close")
elif a==1. and b==1. and c==0:
print("Open")
elif a==0. and b==0. and c==1:
print("Open")
else:
print("Close")
``` | instruction | 0 | 52,961 | 1 | 105,922 |
Yes | output | 1 | 52,961 | 1 | 105,923 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To get on the Shinkansen, you need two tickets, a "ticket" and a "limited express ticket". These are separate tickets because some of the routes may not use the Shinkansen, but for routes that use only the Shinkansen, one ticket can be used as both a ticket and a limited express ticket. It may also be issued.
Automatic ticket gates must read these tickets and open the gate only when the correct ticket is inserted. Create a program that determines whether one "ticket" and one "limited express ticket" or both, or one "boarding / limited express ticket" has been inserted, and determines whether the door of the automatic ticket gate is open or closed. please do it.
input
The input is given in the following format.
b1 b2 b3
The input consists of one line and contains three integers separated by one space. b1 indicates the state in which the "ticket" is inserted, b2 indicates the state in which the "limited express ticket" is inserted, and b3 indicates the state in which the "boarding / limited express ticket" is inserted. The turned-in state is represented by 0 or 1, where 0 indicates a non-charged state, and 1 indicates a loaded state. However, the combination of expected input states is limited to the following cases.
Input | Input state | Door operation for input
--- | --- | ---
1 0 0 | Insert only "ticket" | Close
0 1 0 | Only "Limited Express Ticket" is introduced | Close
1 1 0 | Introducing "tickets" and "express tickets" | Open
0 0 1 | Introducing "Boarding / Limited Express Tickets" | Open
0 0 0 | No input | Close
output
Outputs Open or Close, which indicates the opening and closing of the automatic ticket gate, on one line.
Example
Input
0 0 1
Output
Open
Submitted Solution:
```
a,b,c=map(int,input().split())
if a==1 and b==1 and c==0:
print('Open')
elif a==0 and b==0 and c==1:
print('Open')
else:
print('Close')
``` | instruction | 0 | 52,962 | 1 | 105,924 |
Yes | output | 1 | 52,962 | 1 | 105,925 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To get on the Shinkansen, you need two tickets, a "ticket" and a "limited express ticket". These are separate tickets because some of the routes may not use the Shinkansen, but for routes that use only the Shinkansen, one ticket can be used as both a ticket and a limited express ticket. It may also be issued.
Automatic ticket gates must read these tickets and open the gate only when the correct ticket is inserted. Create a program that determines whether one "ticket" and one "limited express ticket" or both, or one "boarding / limited express ticket" has been inserted, and determines whether the door of the automatic ticket gate is open or closed. please do it.
input
The input is given in the following format.
b1 b2 b3
The input consists of one line and contains three integers separated by one space. b1 indicates the state in which the "ticket" is inserted, b2 indicates the state in which the "limited express ticket" is inserted, and b3 indicates the state in which the "boarding / limited express ticket" is inserted. The turned-in state is represented by 0 or 1, where 0 indicates a non-charged state, and 1 indicates a loaded state. However, the combination of expected input states is limited to the following cases.
Input | Input state | Door operation for input
--- | --- | ---
1 0 0 | Insert only "ticket" | Close
0 1 0 | Only "Limited Express Ticket" is introduced | Close
1 1 0 | Introducing "tickets" and "express tickets" | Open
0 0 1 | Introducing "Boarding / Limited Express Tickets" | Open
0 0 0 | No input | Close
output
Outputs Open or Close, which indicates the opening and closing of the automatic ticket gate, on one line.
Example
Input
0 0 1
Output
Open
Submitted Solution:
```
a,b,c=map(int,input().split())
if a+b+c==1:
print("Close")
elif a+b+c==2:
print("Open")
else:
print("Close")
``` | instruction | 0 | 52,963 | 1 | 105,926 |
No | output | 1 | 52,963 | 1 | 105,927 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To get on the Shinkansen, you need two tickets, a "ticket" and a "limited express ticket". These are separate tickets because some of the routes may not use the Shinkansen, but for routes that use only the Shinkansen, one ticket can be used as both a ticket and a limited express ticket. It may also be issued.
Automatic ticket gates must read these tickets and open the gate only when the correct ticket is inserted. Create a program that determines whether one "ticket" and one "limited express ticket" or both, or one "boarding / limited express ticket" has been inserted, and determines whether the door of the automatic ticket gate is open or closed. please do it.
input
The input is given in the following format.
b1 b2 b3
The input consists of one line and contains three integers separated by one space. b1 indicates the state in which the "ticket" is inserted, b2 indicates the state in which the "limited express ticket" is inserted, and b3 indicates the state in which the "boarding / limited express ticket" is inserted. The turned-in state is represented by 0 or 1, where 0 indicates a non-charged state, and 1 indicates a loaded state. However, the combination of expected input states is limited to the following cases.
Input | Input state | Door operation for input
--- | --- | ---
1 0 0 | Insert only "ticket" | Close
0 1 0 | Only "Limited Express Ticket" is introduced | Close
1 1 0 | Introducing "tickets" and "express tickets" | Open
0 0 1 | Introducing "Boarding / Limited Express Tickets" | Open
0 0 0 | No input | Close
output
Outputs Open or Close, which indicates the opening and closing of the automatic ticket gate, on one line.
Example
Input
0 0 1
Output
Open
Submitted Solution:
```
a,b,c=map(int,input().split())
if a==1:
print("Close")
elif b==1:
print('Close')
elif a+b==2 or c==1:
print("Open")
else:
print("Close")
``` | instruction | 0 | 52,964 | 1 | 105,928 |
No | output | 1 | 52,964 | 1 | 105,929 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To get on the Shinkansen, you need two tickets, a "ticket" and a "limited express ticket". These are separate tickets because some of the routes may not use the Shinkansen, but for routes that use only the Shinkansen, one ticket can be used as both a ticket and a limited express ticket. It may also be issued.
Automatic ticket gates must read these tickets and open the gate only when the correct ticket is inserted. Create a program that determines whether one "ticket" and one "limited express ticket" or both, or one "boarding / limited express ticket" has been inserted, and determines whether the door of the automatic ticket gate is open or closed. please do it.
input
The input is given in the following format.
b1 b2 b3
The input consists of one line and contains three integers separated by one space. b1 indicates the state in which the "ticket" is inserted, b2 indicates the state in which the "limited express ticket" is inserted, and b3 indicates the state in which the "boarding / limited express ticket" is inserted. The turned-in state is represented by 0 or 1, where 0 indicates a non-charged state, and 1 indicates a loaded state. However, the combination of expected input states is limited to the following cases.
Input | Input state | Door operation for input
--- | --- | ---
1 0 0 | Insert only "ticket" | Close
0 1 0 | Only "Limited Express Ticket" is introduced | Close
1 1 0 | Introducing "tickets" and "express tickets" | Open
0 0 1 | Introducing "Boarding / Limited Express Tickets" | Open
0 0 0 | No input | Close
output
Outputs Open or Close, which indicates the opening and closing of the automatic ticket gate, on one line.
Example
Input
0 0 1
Output
Open
Submitted Solution:
```
if __name__ == "__main__":
ticket = input()
reult = "Close"
if ticket in {"1 1 0", "0 0 1"}:
result = "Open"
print(result)
``` | instruction | 0 | 52,965 | 1 | 105,930 |
No | output | 1 | 52,965 | 1 | 105,931 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bessie is out grazing on the farm, which consists of n fields connected by m bidirectional roads. She is currently at field 1, and will return to her home at field n at the end of the day.
The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has k special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.
After the road is added, Bessie will return home on the shortest path from field 1 to field n. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!
Input
The first line contains integers n, m, and k (2 ≤ n ≤ 2 ⋅ 10^5, n-1 ≤ m ≤ 2 ⋅ 10^5, 2 ≤ k ≤ n) — the number of fields on the farm, the number of roads, and the number of special fields.
The second line contains k integers a_1, a_2, …, a_k (1 ≤ a_i ≤ n) — the special fields. All a_i are distinct.
The i-th of the following m lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), representing a bidirectional road between fields x_i and y_i.
It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.
Output
Output one integer, the maximum possible length of the shortest path from field 1 to n after Farmer John installs one road optimally.
Examples
Input
5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
Output
3
Input
5 4 2
2 4
1 2
2 3
3 4
4 5
Output
3
Note
The graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields 3 and 5, and the resulting shortest path from 1 to 5 is length 3.
<image>
The graph for the second example is shown below. Farmer John must add a road between fields 2 and 4, and the resulting shortest path from 1 to 5 is length 3.
<image> | instruction | 0 | 53,202 | 1 | 106,404 |
Tags: binary search, data structures, dfs and similar, graphs, greedy, shortest paths, sortings
Correct Solution:
```
import sys
from collections import deque
input=sys.stdin.readline
n,m,k=map(int,input().split())
a=list(map(int,input().split()))
a=[a[i]-1 for i in range(k)]
edge=[[] for i in range(n)]
for i in range(m):
x,y=map(int,input().split())
edge[x-1].append(y-1)
edge[y-1].append(x-1)
def bfs(s):
dist=[10**20]*n
que=deque([s])
dist[s]=0
while que:
v=que.popleft()
for nv in edge[v]:
if dist[nv]>dist[v]+1:
dist[nv]=dist[v]+1
que.append(nv)
return dist
data1=bfs(0)
data2=bfs(n-1)
s=[]
e=[]
for i in range(k):
s.append(data1[a[i]])
e.append(data2[a[i]])
S=[]
E=[]
for i in range(k):
S.append((e[i]-s[i],i,s[i]))
E.append((e[i]-s[i],i,e[i]))
S.sort()
E.sort()
EE=[E[i][2] for i in range(k)]
for i in range(1,k):
EE[i]=max(EE[i],EE[i-1])
ans=0
for i in range(1,k):
test=S[i][2]+EE[i-1]+1
ans=max(ans,test)
print(min(data1[-1],ans))
``` | output | 1 | 53,202 | 1 | 106,405 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bessie is out grazing on the farm, which consists of n fields connected by m bidirectional roads. She is currently at field 1, and will return to her home at field n at the end of the day.
The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has k special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.
After the road is added, Bessie will return home on the shortest path from field 1 to field n. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!
Input
The first line contains integers n, m, and k (2 ≤ n ≤ 2 ⋅ 10^5, n-1 ≤ m ≤ 2 ⋅ 10^5, 2 ≤ k ≤ n) — the number of fields on the farm, the number of roads, and the number of special fields.
The second line contains k integers a_1, a_2, …, a_k (1 ≤ a_i ≤ n) — the special fields. All a_i are distinct.
The i-th of the following m lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), representing a bidirectional road between fields x_i and y_i.
It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.
Output
Output one integer, the maximum possible length of the shortest path from field 1 to n after Farmer John installs one road optimally.
Examples
Input
5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
Output
3
Input
5 4 2
2 4
1 2
2 3
3 4
4 5
Output
3
Note
The graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields 3 and 5, and the resulting shortest path from 1 to 5 is length 3.
<image>
The graph for the second example is shown below. Farmer John must add a road between fields 2 and 4, and the resulting shortest path from 1 to 5 is length 3.
<image> | instruction | 0 | 53,203 | 1 | 106,406 |
Tags: binary search, data structures, dfs and similar, graphs, greedy, shortest paths, sortings
Correct Solution:
```
def bfs(V):
q = deque()
q.append(V)
distance = [-1] * n
distance[V]=0
while len(q) > 0 :
v = q.popleft()
for u in g[v]:
if distance[u]!=-1:
continue
distance[u] = distance[v] + 1
q.append(u)
return distance
from collections import deque
import sys
n,m,k=map(int,input().split())
a=list(map(lambda x:x-1,map(int,input().split())))
input=sys.stdin.readlines()
g=[[] for i in range(n)]
for i in range(m):
x,y=map(int,input[i].split())
x-=1
y-=1
g[x].append(y)
g[y].append(x)
dstart=bfs(0)
dend=bfs(n-1)
p=sorted([(dstart[i],dend[i]) for i in a])
maxs=p[0][0]
answer=0
for i in range(1,k):
answer=max(maxs+p[i][1]+1,answer)
maxs=max(maxs,p[i][0])
print(min(answer,dend[0]))
``` | output | 1 | 53,203 | 1 | 106,407 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bessie is out grazing on the farm, which consists of n fields connected by m bidirectional roads. She is currently at field 1, and will return to her home at field n at the end of the day.
The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has k special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.
After the road is added, Bessie will return home on the shortest path from field 1 to field n. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!
Input
The first line contains integers n, m, and k (2 ≤ n ≤ 2 ⋅ 10^5, n-1 ≤ m ≤ 2 ⋅ 10^5, 2 ≤ k ≤ n) — the number of fields on the farm, the number of roads, and the number of special fields.
The second line contains k integers a_1, a_2, …, a_k (1 ≤ a_i ≤ n) — the special fields. All a_i are distinct.
The i-th of the following m lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), representing a bidirectional road between fields x_i and y_i.
It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.
Output
Output one integer, the maximum possible length of the shortest path from field 1 to n after Farmer John installs one road optimally.
Examples
Input
5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
Output
3
Input
5 4 2
2 4
1 2
2 3
3 4
4 5
Output
3
Note
The graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields 3 and 5, and the resulting shortest path from 1 to 5 is length 3.
<image>
The graph for the second example is shown below. Farmer John must add a road between fields 2 and 4, and the resulting shortest path from 1 to 5 is length 3.
<image> | instruction | 0 | 53,204 | 1 | 106,408 |
Tags: binary search, data structures, dfs and similar, graphs, greedy, shortest paths, sortings
Correct Solution:
```
from math import factorial
from collections import Counter, defaultdict, deque
from heapq import heapify, heappop, heappush
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def comb(n, m): return factorial(n) / (factorial(m) * factorial(n - m)) if n >= m else 0
def perm(n, m): return factorial(n) // (factorial(n - m)) if n >= m else 0
def mdis(x1, y1, x2, y2): return abs(x1 - x2) + abs(y1 - y2)
mod = 1000000007
INF = float('inf')
# ------------------------------
def main():
n, m, k = RL()
spa = list(RL())
dic = defaultdict(set)
ds = 0
for _ in range(m):
a, b = RL()
dic[a].add(b)
dic[b].add(a)
def bfs(root):
rec = [0]*(n+1)
q = [(root, 0)]
da = [0]*(n+1)
rec[root] = 1
for i, d in q:
for nex in dic[i]:
if rec[nex]==0:
q.append((nex, d+1))
rec[nex] = 1
da[nex] = d+1
return da
star = bfs(1)
end = bfs(n)
res = end[1]
spa.sort(key=lambda i: star[i]-end[i])
ts = float('-inf')
disa = star[spa[0]]
dis = 0
# print(end, spa)
for i in spa[1:]:
dis = max(end[i]+1+disa, dis)
disa = max(disa, star[i])
# print(dis, disa, i)
print(min(res, dis))
if __name__ == "__main__":
main()
``` | output | 1 | 53,204 | 1 | 106,409 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bessie is out grazing on the farm, which consists of n fields connected by m bidirectional roads. She is currently at field 1, and will return to her home at field n at the end of the day.
The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has k special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.
After the road is added, Bessie will return home on the shortest path from field 1 to field n. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!
Input
The first line contains integers n, m, and k (2 ≤ n ≤ 2 ⋅ 10^5, n-1 ≤ m ≤ 2 ⋅ 10^5, 2 ≤ k ≤ n) — the number of fields on the farm, the number of roads, and the number of special fields.
The second line contains k integers a_1, a_2, …, a_k (1 ≤ a_i ≤ n) — the special fields. All a_i are distinct.
The i-th of the following m lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), representing a bidirectional road between fields x_i and y_i.
It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.
Output
Output one integer, the maximum possible length of the shortest path from field 1 to n after Farmer John installs one road optimally.
Examples
Input
5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
Output
3
Input
5 4 2
2 4
1 2
2 3
3 4
4 5
Output
3
Note
The graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields 3 and 5, and the resulting shortest path from 1 to 5 is length 3.
<image>
The graph for the second example is shown below. Farmer John must add a road between fields 2 and 4, and the resulting shortest path from 1 to 5 is length 3.
<image> | instruction | 0 | 53,205 | 1 | 106,410 |
Tags: binary search, data structures, dfs and similar, graphs, greedy, shortest paths, sortings
Correct Solution:
```
from collections import deque
import sys
input=sys.stdin.readline
n,m,k=map(int,input().split())
A=list(map(int,input().split()))
E=[[] for _ in range(n)]
for _ in range(m):
x,y=map(int,input().split())
E[x-1].append(y-1)
E[y-1].append(x-1)
d=[[-1,-1] for _ in range(n)]
d[0][0]=0
todo=deque([0])
while todo:
v=todo.popleft()
for u in E[v]:
if d[u][0]==-1:
todo.append(u)
d[u][0]=d[v][0]+1
d[n-1][1]=0
todo=deque([n-1])
while todo:
v=todo.popleft()
for u in E[v]:
if d[u][1]==-1:
todo.append(u)
d[u][1]=d[v][1]+1
ans=d[n-1][0]
d_=[]
for a in A:
d_.append(d[a-1])
d_.sort()
d_g=[0]*k
d_g[k-1]=d_[k-1][1]
for i in range(k-1,0,-1):
d_g[i-1]=max(d_g[i],d_[i-1][1])
ans_=0
for i in range(k-1):
ans_=max(ans_,d_[i][0]+d_g[i+1]+1)
print(min(ans,ans_))
``` | output | 1 | 53,205 | 1 | 106,411 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bessie is out grazing on the farm, which consists of n fields connected by m bidirectional roads. She is currently at field 1, and will return to her home at field n at the end of the day.
The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has k special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.
After the road is added, Bessie will return home on the shortest path from field 1 to field n. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!
Input
The first line contains integers n, m, and k (2 ≤ n ≤ 2 ⋅ 10^5, n-1 ≤ m ≤ 2 ⋅ 10^5, 2 ≤ k ≤ n) — the number of fields on the farm, the number of roads, and the number of special fields.
The second line contains k integers a_1, a_2, …, a_k (1 ≤ a_i ≤ n) — the special fields. All a_i are distinct.
The i-th of the following m lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), representing a bidirectional road between fields x_i and y_i.
It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.
Output
Output one integer, the maximum possible length of the shortest path from field 1 to n after Farmer John installs one road optimally.
Examples
Input
5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
Output
3
Input
5 4 2
2 4
1 2
2 3
3 4
4 5
Output
3
Note
The graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields 3 and 5, and the resulting shortest path from 1 to 5 is length 3.
<image>
The graph for the second example is shown below. Farmer John must add a road between fields 2 and 4, and the resulting shortest path from 1 to 5 is length 3.
<image> | instruction | 0 | 53,206 | 1 | 106,412 |
Tags: binary search, data structures, dfs and similar, graphs, greedy, shortest paths, sortings
Correct Solution:
```
import sys
input = sys.stdin.readline
n,m,K = map(int,input().split())
sp = list(map(lambda x:int(x)-1 ,input().split()))
peer = [[] for _ in range(n)]
for _ in range(m):
a,b = map(int,input().split())
a -= 1
b -= 1
peer[a].append(b)
peer[b].append(a)
a0 = [10 ** 6 for _ in range(n)]
a0[0] = 0
a1 = [10 ** 6 for _ in range(n)]
a1[-1] = 0
now = [0]
while now:
last = now
now = []
for x in last:
for y in peer[x]:
if a0[y] > a0[x] + 1:
a0[y] = a0[x] + 1
now.append(y)
#print(a0)
now = [n-1]
while now:
last = now
now = []
for x in last:
for y in peer[x]:
if a1[y] > a1[x] + 1:
a1[y] = a1[x] + 1
now.append(y)
xyxy = []
for w in sp:
xyxy.append([a0[w]-a1[w],a0[w],a1[w]])
xyxy.sort()
#print(xyxy)
xx = []
maxi = 0
for j in range(K):
xx.append(max(maxi,xyxy[j][1]))
#print(xx)
can = []
for j in range(1,K):
can.append(xx[j-1]+xyxy[j][2]+1)
print(min(a0[-1],max(can)))
#print(can)
``` | output | 1 | 53,206 | 1 | 106,413 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bessie is out grazing on the farm, which consists of n fields connected by m bidirectional roads. She is currently at field 1, and will return to her home at field n at the end of the day.
The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has k special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.
After the road is added, Bessie will return home on the shortest path from field 1 to field n. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!
Input
The first line contains integers n, m, and k (2 ≤ n ≤ 2 ⋅ 10^5, n-1 ≤ m ≤ 2 ⋅ 10^5, 2 ≤ k ≤ n) — the number of fields on the farm, the number of roads, and the number of special fields.
The second line contains k integers a_1, a_2, …, a_k (1 ≤ a_i ≤ n) — the special fields. All a_i are distinct.
The i-th of the following m lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), representing a bidirectional road between fields x_i and y_i.
It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.
Output
Output one integer, the maximum possible length of the shortest path from field 1 to n after Farmer John installs one road optimally.
Examples
Input
5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
Output
3
Input
5 4 2
2 4
1 2
2 3
3 4
4 5
Output
3
Note
The graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields 3 and 5, and the resulting shortest path from 1 to 5 is length 3.
<image>
The graph for the second example is shown below. Farmer John must add a road between fields 2 and 4, and the resulting shortest path from 1 to 5 is length 3.
<image> | instruction | 0 | 53,207 | 1 | 106,414 |
Tags: binary search, data structures, dfs and similar, graphs, greedy, shortest paths, sortings
Correct Solution:
```
#!/usr/bin/env python3
import sys
input = sys.stdin.readline
import heapq
from heapq import heappop, heappush
INF = 10**9
class Dijkstra:
def __init__(self, adj):
self.n = len(adj)
self.adj = adj
self.dist = [INF] * len(adj)
self.q = []
def reset(self):
self.dist = [INF] * self.n
self.q = []
def calc(self, start):
self.dist[start] = 0
heapq.heappush(self.q, (0, start))
while len(self.q) != 0:
prov_cost, src = heapq.heappop(self.q)
if self.dist[src] < prov_cost:
continue
for dest in self.adj[src]:
if self.dist[dest] > self.dist[src] + 1:
self.dist[dest] = self.dist[src] + 1
heapq.heappush(self.q, (self.dist[dest], dest))
return self.dist
n, m, k = map(int, input().split())
special = set([int(item) - 1 for item in input().split()])
edge = [[] for _ in range(n)]
for _ in range(m):
u, v = map(int, input().split())
u -= 1; v -= 1
edge[u].append(v)
edge[v].append(u)
DIJK = Dijkstra(edge)
forward = DIJK.calc(0)
DIJK.reset()
backward = DIJK.calc(n-1)
sp = []
for i, (ff, bb) in enumerate(zip(forward, backward)):
if i not in special:
continue
sp.append((ff, bb))
sp.sort(reverse=True)
queue = []
current = sp[1][0]
heapq.heappush(queue, -sp[0][1])
heapq.heappush(queue, -sp[1][1])
ans = current + min(sp[0][1], sp[1][1])
for ff, bb in sp[2:]:
current = ff
heappush(queue, -bb)
m1 = heappop(queue)
m2 = heappop(queue)
ans = max(ans, current - max(m1, m2))
heappush(queue, m1)
heappush(queue, m2)
print(min(ans + 1, forward[n-1]))
``` | output | 1 | 53,207 | 1 | 106,415 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bessie is out grazing on the farm, which consists of n fields connected by m bidirectional roads. She is currently at field 1, and will return to her home at field n at the end of the day.
The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has k special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.
After the road is added, Bessie will return home on the shortest path from field 1 to field n. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!
Input
The first line contains integers n, m, and k (2 ≤ n ≤ 2 ⋅ 10^5, n-1 ≤ m ≤ 2 ⋅ 10^5, 2 ≤ k ≤ n) — the number of fields on the farm, the number of roads, and the number of special fields.
The second line contains k integers a_1, a_2, …, a_k (1 ≤ a_i ≤ n) — the special fields. All a_i are distinct.
The i-th of the following m lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), representing a bidirectional road between fields x_i and y_i.
It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.
Output
Output one integer, the maximum possible length of the shortest path from field 1 to n after Farmer John installs one road optimally.
Examples
Input
5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
Output
3
Input
5 4 2
2 4
1 2
2 3
3 4
4 5
Output
3
Note
The graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields 3 and 5, and the resulting shortest path from 1 to 5 is length 3.
<image>
The graph for the second example is shown below. Farmer John must add a road between fields 2 and 4, and the resulting shortest path from 1 to 5 is length 3.
<image> | instruction | 0 | 53,208 | 1 | 106,416 |
Tags: binary search, data structures, dfs and similar, graphs, greedy, shortest paths, sortings
Correct Solution:
```
#Pye
from sys import stdin, stdout
from os import path
from collections import deque
from heapq import heappop, heappush
maxn = 200005
mark = [0 for i in range(maxn)]
f = [0 for i in range(maxn)]
g = [0 for i in range(maxn)]
a = [[] for i in range(maxn)]
ans = 0
def bfs(s, f):
global a
q = deque()
q.append(s)
while len(q) > 0:
u = q.popleft()
for v in a[u]:
if v == s: continue
if not f[v]:
f[v] = f[u] + 1
q.append(v)
if path.exists('inp.txt'):
stdin = open("inp.txt", "r")
n, m, k = map(int, stdin.readline().split())
p = list(map(int, stdin.readline().split()))
for i in p: mark[i] = 1
for i in range(m):
u, v = map(int, stdin.readline().split())
a[u].append(v)
a[v].append(u)
bfs(1, f)
bfs(n, g)
node = []
for i in range(1, n+1):
if mark[i]: node.append((f[i], g[i]))
node.sort()
pq = []
for i in range(len(node)): heappush(pq, (-node[i][1], i))
for i in range(len(node)):
while len(pq) > 0:
u = heappop(pq)
if u[1] <= i: continue
else:
heappush(pq, u)
break
if not len(pq): continue
tmp = heappop(pq); heappush(pq, tmp)
u = node[i]; v = node[tmp[1]]
res = min(u[0] + v[1], u[1] + v[0]) + 1
res = min(res, min(u[0] + u[1], v[0] + v[1]))
ans = max(ans, res)
stdout.write(str(min(ans, f[n])))
``` | output | 1 | 53,208 | 1 | 106,417 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bessie is out grazing on the farm, which consists of n fields connected by m bidirectional roads. She is currently at field 1, and will return to her home at field n at the end of the day.
The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has k special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.
After the road is added, Bessie will return home on the shortest path from field 1 to field n. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!
Input
The first line contains integers n, m, and k (2 ≤ n ≤ 2 ⋅ 10^5, n-1 ≤ m ≤ 2 ⋅ 10^5, 2 ≤ k ≤ n) — the number of fields on the farm, the number of roads, and the number of special fields.
The second line contains k integers a_1, a_2, …, a_k (1 ≤ a_i ≤ n) — the special fields. All a_i are distinct.
The i-th of the following m lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), representing a bidirectional road between fields x_i and y_i.
It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.
Output
Output one integer, the maximum possible length of the shortest path from field 1 to n after Farmer John installs one road optimally.
Examples
Input
5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
Output
3
Input
5 4 2
2 4
1 2
2 3
3 4
4 5
Output
3
Note
The graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields 3 and 5, and the resulting shortest path from 1 to 5 is length 3.
<image>
The graph for the second example is shown below. Farmer John must add a road between fields 2 and 4, and the resulting shortest path from 1 to 5 is length 3.
<image> | instruction | 0 | 53,209 | 1 | 106,418 |
Tags: binary search, data structures, dfs and similar, graphs, greedy, shortest paths, sortings
Correct Solution:
```
def bfs(adj, root):
q = [(root, 0)]
dist = [0] * len(adj)
vis = {root}
i = 0
while i < len(q):
u, d = q[i]
i += 1
dist[u] = d
for v in adj[u]:
if v not in vis:
vis.add(v)
q.append((v, d + 1))
return dist
def sol(n, sfs, edges):
adj = [[] for _ in range(n + 1)]
for u, v in edges:
adj[u].append(v)
adj[v].append(u)
dista = bfs(adj, 1)
distb = bfs(adj, n)
orig = distb[1]
ans = 0
sfs.sort(key=lambda sf: dista[sf] - distb[sf])
maxdista = dista[sfs[0]]
for fs in sfs[1:]:
ans = max(ans, maxdista + distb[fs])
maxdista = max(maxdista, dista[fs])
return min(orig, ans + 1)
n, m, _ = map(int, input().split())
sfs = list(map(int, input().split()))
edges = [tuple(map(int, input().split())) for _ in range(m)]
print(sol(n, sfs, edges))
``` | output | 1 | 53,209 | 1 | 106,419 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bessie is out grazing on the farm, which consists of n fields connected by m bidirectional roads. She is currently at field 1, and will return to her home at field n at the end of the day.
The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has k special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.
After the road is added, Bessie will return home on the shortest path from field 1 to field n. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!
Input
The first line contains integers n, m, and k (2 ≤ n ≤ 2 ⋅ 10^5, n-1 ≤ m ≤ 2 ⋅ 10^5, 2 ≤ k ≤ n) — the number of fields on the farm, the number of roads, and the number of special fields.
The second line contains k integers a_1, a_2, …, a_k (1 ≤ a_i ≤ n) — the special fields. All a_i are distinct.
The i-th of the following m lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), representing a bidirectional road between fields x_i and y_i.
It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.
Output
Output one integer, the maximum possible length of the shortest path from field 1 to n after Farmer John installs one road optimally.
Examples
Input
5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
Output
3
Input
5 4 2
2 4
1 2
2 3
3 4
4 5
Output
3
Note
The graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields 3 and 5, and the resulting shortest path from 1 to 5 is length 3.
<image>
The graph for the second example is shown below. Farmer John must add a road between fields 2 and 4, and the resulting shortest path from 1 to 5 is length 3.
<image>
Submitted Solution:
```
from math import *
from collections import deque
INF = (1<<60)
def main():
n,m,k = map(int,input().split(' '))
spec = set(map(int,input().split(' ')))
adjl = [[] for _ in range(0,n+1)]
for i in range(0,m):
u,v = map(int,input().split(' '))
adjl[u].append(v)
adjl[v].append(u)
srcdst = bfs(n,adjl,1)
enddst = bfs(n,adjl,n)
diff = [[srcdst[i]-enddst[i],i] for i in range(1,n+1) if i in spec]
diff.sort(key = lambda x:x[0])
mxsrc = -INF
res = 0
for x in range(0,len(diff)):
res = max(res, mxsrc + enddst[diff[x][1]] + 1)
mxsrc = max(mxsrc, srcdst[diff[x][1]])
print(min(srcdst[n],res))
def bfs(n,adjl,src):
q = deque()
q.append(src)
odist = [INF for _ in range(0,n+1)]
vis = [False for _ in range(0,n+1)]
odist[src] = 0
vis[src] = True
while(len(q) > 0):
cur = q.popleft()
for adj in adjl[cur]:
if not vis[adj]:
vis[adj] = True
odist[adj] = odist[cur]+1
q.append(adj)
return odist
if __name__ == '__main__':
main()
``` | instruction | 0 | 53,210 | 1 | 106,420 |
Yes | output | 1 | 53,210 | 1 | 106,421 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bessie is out grazing on the farm, which consists of n fields connected by m bidirectional roads. She is currently at field 1, and will return to her home at field n at the end of the day.
The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has k special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.
After the road is added, Bessie will return home on the shortest path from field 1 to field n. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!
Input
The first line contains integers n, m, and k (2 ≤ n ≤ 2 ⋅ 10^5, n-1 ≤ m ≤ 2 ⋅ 10^5, 2 ≤ k ≤ n) — the number of fields on the farm, the number of roads, and the number of special fields.
The second line contains k integers a_1, a_2, …, a_k (1 ≤ a_i ≤ n) — the special fields. All a_i are distinct.
The i-th of the following m lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), representing a bidirectional road between fields x_i and y_i.
It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.
Output
Output one integer, the maximum possible length of the shortest path from field 1 to n after Farmer John installs one road optimally.
Examples
Input
5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
Output
3
Input
5 4 2
2 4
1 2
2 3
3 4
4 5
Output
3
Note
The graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields 3 and 5, and the resulting shortest path from 1 to 5 is length 3.
<image>
The graph for the second example is shown below. Farmer John must add a road between fields 2 and 4, and the resulting shortest path from 1 to 5 is length 3.
<image>
Submitted Solution:
```
import sys
input = sys.stdin.readline
N, M, K = map(int, input().split())
A = list(map(lambda x: int(x)-1, input().split()))
graph = [[] for _ in range(N)]
for _ in range(M):
a, b = map(int, input().split())
graph[a-1].append(b-1)
graph[b-1].append(a-1)
def bfs(s):
D = [-1]*N
D[s] = 0
q = [s]
while q:
qq = []
for p in q:
for np in graph[p]:
if D[np] == -1:
D[np] = D[p] + 1
qq.append(np)
q = qq
return D
D1 = bfs(0)
D2 = bfs(N-1)
distance = D1[N-1]
P = []
for a in A:
d1 = D1[a]
d2 = D2[a]
P.append((d1, d2))
P.sort()
ans = 0
for i in range(K-1):
b1, b2 = P[i]
c1, c2 = P[i+1]
ans = max(ans, min(c1+b2+1, c2+b1+1))
ans = min(ans, distance)
print(ans)
``` | instruction | 0 | 53,211 | 1 | 106,422 |
Yes | output | 1 | 53,211 | 1 | 106,423 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bessie is out grazing on the farm, which consists of n fields connected by m bidirectional roads. She is currently at field 1, and will return to her home at field n at the end of the day.
The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has k special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.
After the road is added, Bessie will return home on the shortest path from field 1 to field n. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!
Input
The first line contains integers n, m, and k (2 ≤ n ≤ 2 ⋅ 10^5, n-1 ≤ m ≤ 2 ⋅ 10^5, 2 ≤ k ≤ n) — the number of fields on the farm, the number of roads, and the number of special fields.
The second line contains k integers a_1, a_2, …, a_k (1 ≤ a_i ≤ n) — the special fields. All a_i are distinct.
The i-th of the following m lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), representing a bidirectional road between fields x_i and y_i.
It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.
Output
Output one integer, the maximum possible length of the shortest path from field 1 to n after Farmer John installs one road optimally.
Examples
Input
5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
Output
3
Input
5 4 2
2 4
1 2
2 3
3 4
4 5
Output
3
Note
The graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields 3 and 5, and the resulting shortest path from 1 to 5 is length 3.
<image>
The graph for the second example is shown below. Farmer John must add a road between fields 2 and 4, and the resulting shortest path from 1 to 5 is length 3.
<image>
Submitted Solution:
```
# -*- coding: utf-8 -*-
import math
import collections
import bisect
import heapq
import time
import random
import itertools
import sys
from typing import List
"""
created by shhuan at 2020/3/14 13:31
"""
def bfs(s, t, edges, dist):
dist[t] = 0
step = 0
q = [t]
while q:
step += 1
nq = []
for u in q:
for v in edges[u]:
if step < dist[v]:
dist[v] = step
nq.append(v)
q = nq
def solve(N, M, K, A, edges):
dista = [N+1 for _ in range(N+1)]
distb = [N+1 for _ in range(N+1)]
bfs(1, N, edges, dista)
bfs(N, 1, edges, distb)
xyd = [(dista[i]-distb[i], i) for i in range(1, N+1) if A[i]]
xyd.sort()
maxx = float('-inf')
best = 0
for d, i in xyd:
best = max(best, distb[i] + maxx)
maxx = max(maxx, dista[i])
return min(dista[1], best + 1)
N, M, K = map(int, input().split())
specials = [int(x) for x in input().split()]
A = [False for _ in range(N+1)]
for v in specials:
A[v] = True
edges = [[] for _ in range(N+1)]
for i in range(M):
u, v = map(int, input().split())
edges[u].append(v)
edges[v].append(u)
print(solve(N, M, K, A, edges))
``` | instruction | 0 | 53,212 | 1 | 106,424 |
Yes | output | 1 | 53,212 | 1 | 106,425 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bessie is out grazing on the farm, which consists of n fields connected by m bidirectional roads. She is currently at field 1, and will return to her home at field n at the end of the day.
The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has k special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.
After the road is added, Bessie will return home on the shortest path from field 1 to field n. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!
Input
The first line contains integers n, m, and k (2 ≤ n ≤ 2 ⋅ 10^5, n-1 ≤ m ≤ 2 ⋅ 10^5, 2 ≤ k ≤ n) — the number of fields on the farm, the number of roads, and the number of special fields.
The second line contains k integers a_1, a_2, …, a_k (1 ≤ a_i ≤ n) — the special fields. All a_i are distinct.
The i-th of the following m lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), representing a bidirectional road between fields x_i and y_i.
It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.
Output
Output one integer, the maximum possible length of the shortest path from field 1 to n after Farmer John installs one road optimally.
Examples
Input
5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
Output
3
Input
5 4 2
2 4
1 2
2 3
3 4
4 5
Output
3
Note
The graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields 3 and 5, and the resulting shortest path from 1 to 5 is length 3.
<image>
The graph for the second example is shown below. Farmer John must add a road between fields 2 and 4, and the resulting shortest path from 1 to 5 is length 3.
<image>
Submitted Solution:
```
from heapq import heappush, heappop, heapify
from collections import deque, defaultdict, Counter
import itertools
from itertools import permutations, combinations, accumulate
import sys
import bisect
import string
import math
import time
def I(): return int(input())
def MI(): return map(int, input().split())
def S(): return input()
def MS(): return map(str, input().split())
def LI(): return [int(i) for i in input().split()]
def LI_(): return [int(i)-1 for i in input().split()]
def StoI(): return [ord(i)-97 for i in input()]
def ItoS(nn): return chr(nn+97)
def input(): return sys.stdin.readline().rstrip()
def show(*inp, end='\n'):
if show_flg:
print(*inp, end=end)
YN = {False: 'No', True: 'Yes'}
MOD = 10**9+7
inf = float('inf')
IINF = 10**10
l_alp = string.ascii_lowercase
u_alp = string.ascii_uppercase
nums = ['1', '2', '3', '4', '5', '6', '7', '8', '9', '10']
show_flg = False
# show_flg = True
def bfs(graph, initial, n):
dist = [-1] * (n + 1)
q = deque([initial])
visited = [False] * (n + 1)
visited[initial] = True
dist[initial] = 0
while len(q) != 0:
edge = q.popleft()
nxt = graph[edge]
for i, e in enumerate(nxt):
if visited[e] is False:
q.append(e)
dist[e] = dist[edge] + 1
visited[e] = True
return dist
def main():
n, m, k = MI()
a = LI()
graph = [[] for i in range(n+1)]
is_special = {}
for i in a:
is_special[i] = True
for i in range(m):
x, y = map(int, input().split())
graph[x].append(y)
graph[y].append(x)
dist_1toN = bfs(graph, 1, n)
dist_Nto1 = bfs(graph, n, n)
data = [None] * k
for i in range(len(a)):
x = dist_1toN[a[i]]
y = dist_Nto1[a[i]]
data[i] = (x, y)
data.sort()
# a = [d[1] for d in data]
best = 0
for i in range(len(data)-1):
best = max(best, data[i][0] + 1 + data[i+1][1])
print(min(best, dist_1toN[n]))
# data = [None] * k
# for i in range(len(a)):
# x = dist_1toN[a[i]]
# y = dist_Nto1[a[i]]
# data[i] = (x - y, a[i])
# data.sort()
# a = [d[1] for d in data]
# best = 0
# mx = -10**9
# for i in range(len(a)):
# best = max(best, mx + dist_Nto1[a[i]] + 1)
# mx = max(mx, dist_1toN[a[i]])
# # print(a[i], best, mx)
# print(min(dist_1toN[n], best))
if __name__ == '__main__':
main()
``` | instruction | 0 | 53,213 | 1 | 106,426 |
Yes | output | 1 | 53,213 | 1 | 106,427 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bessie is out grazing on the farm, which consists of n fields connected by m bidirectional roads. She is currently at field 1, and will return to her home at field n at the end of the day.
The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has k special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.
After the road is added, Bessie will return home on the shortest path from field 1 to field n. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!
Input
The first line contains integers n, m, and k (2 ≤ n ≤ 2 ⋅ 10^5, n-1 ≤ m ≤ 2 ⋅ 10^5, 2 ≤ k ≤ n) — the number of fields on the farm, the number of roads, and the number of special fields.
The second line contains k integers a_1, a_2, …, a_k (1 ≤ a_i ≤ n) — the special fields. All a_i are distinct.
The i-th of the following m lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), representing a bidirectional road between fields x_i and y_i.
It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.
Output
Output one integer, the maximum possible length of the shortest path from field 1 to n after Farmer John installs one road optimally.
Examples
Input
5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
Output
3
Input
5 4 2
2 4
1 2
2 3
3 4
4 5
Output
3
Note
The graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields 3 and 5, and the resulting shortest path from 1 to 5 is length 3.
<image>
The graph for the second example is shown below. Farmer John must add a road between fields 2 and 4, and the resulting shortest path from 1 to 5 is length 3.
<image>
Submitted Solution:
```
import sys
input=sys.stdin.readline
import collections
from collections import defaultdict
def bfs(graph, root):
visited=set()
parent=[]
if l1[root]==1:
parent.append([0,root])
dist=[0]*(n+1)
count=0
queue=collections.deque([root])
visited.add(root)
while queue:
vertex=queue.popleft()
for neighbour in graph[vertex]:
if neighbour not in visited:
if l1[neighbour]==1:
parent.append([dist[vertex]+1,neighbour])
dist[neighbour]=dist[vertex]+1
visited.add(neighbour)
queue.append(neighbour)
return(parent,dist)
n,m,k=map(int,input().split())
special=[int(i) for i in input().split()]
l1=[0]*(n+1)
for j in special:
l1[j]+=1
graph=defaultdict(list)
ok=False
for i in range(m):
x,y=map(int,input().split())
if l1[x]==1 and l1[y]==1 or (x==1 and y==n) or (y==n and x==1):
ok=True
graph[x].append(y)
graph[y].append(x)
outta,putta=(bfs(graph,1))
copy=outta[:]
lk,kl=bfs(graph,n)
new=[]
for ljk in range(len(kl)):
if (putta[ljk]!=0 or kl[ljk]!=0) and l1[ljk]==1:
new.append([putta[ljk],kl[ljk]])
#print(new,putta,kl)
new.sort()
for i in range(len(new)-1):
if new[i+1][0]>=new[i][0] and new[i+1][1]>=new[i][1]:
ok=True
break
#print(outta,putta)
if ok==True :
print(putta[n])
else:
min_=1000000
copy.sort()
#print(copy,outta)
for j in range(1,len(new)):
min_=min(min_,new[j][0]-new[j-1][0],new[-j-1][1]-new[-j][1])
print(putta[n]-(min_)+1)
``` | instruction | 0 | 53,214 | 1 | 106,428 |
No | output | 1 | 53,214 | 1 | 106,429 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bessie is out grazing on the farm, which consists of n fields connected by m bidirectional roads. She is currently at field 1, and will return to her home at field n at the end of the day.
The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has k special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.
After the road is added, Bessie will return home on the shortest path from field 1 to field n. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!
Input
The first line contains integers n, m, and k (2 ≤ n ≤ 2 ⋅ 10^5, n-1 ≤ m ≤ 2 ⋅ 10^5, 2 ≤ k ≤ n) — the number of fields on the farm, the number of roads, and the number of special fields.
The second line contains k integers a_1, a_2, …, a_k (1 ≤ a_i ≤ n) — the special fields. All a_i are distinct.
The i-th of the following m lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), representing a bidirectional road between fields x_i and y_i.
It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.
Output
Output one integer, the maximum possible length of the shortest path from field 1 to n after Farmer John installs one road optimally.
Examples
Input
5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
Output
3
Input
5 4 2
2 4
1 2
2 3
3 4
4 5
Output
3
Note
The graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields 3 and 5, and the resulting shortest path from 1 to 5 is length 3.
<image>
The graph for the second example is shown below. Farmer John must add a road between fields 2 and 4, and the resulting shortest path from 1 to 5 is length 3.
<image>
Submitted Solution:
```
def bfs(adj, root):
q = [(root, 0)]
dist = [0] * len(adj)
vis = {root}
i = 0
while i < len(q):
u, d = q[i]
i += 1
dist[u] = d
for v in adj[u]:
if v not in vis:
vis.add(v)
q.append((v, d + 1))
return dist
def sol(n, sfs, edges):
adj = [[] for _ in range(n + 1)]
for u, v in edges:
adj[u].append(v)
adj[v].append(u)
dista = bfs(adj, 1)
distb = bfs(adj, n)
orig = distb[1]
ans = 0
distsfs = sorted((sf, dista[sf] - distb[sf]) for sf in sfs)
maxdista = dista[distsfs[0][0]]
for fs, _ in distsfs[1:]:
ans = max(ans, maxdista + distb[fs] + 1)
if ans >= orig:
break
maxdista = max(maxdista, dista[fs])
return min(orig, ans)
n, m, _ = map(int, input().split())
sfs = list(map(int, input().split()))
edges = [tuple(map(int, input().split())) for _ in range(m)]
print(sol(n, sfs, edges))
``` | instruction | 0 | 53,215 | 1 | 106,430 |
No | output | 1 | 53,215 | 1 | 106,431 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bessie is out grazing on the farm, which consists of n fields connected by m bidirectional roads. She is currently at field 1, and will return to her home at field n at the end of the day.
The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has k special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.
After the road is added, Bessie will return home on the shortest path from field 1 to field n. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!
Input
The first line contains integers n, m, and k (2 ≤ n ≤ 2 ⋅ 10^5, n-1 ≤ m ≤ 2 ⋅ 10^5, 2 ≤ k ≤ n) — the number of fields on the farm, the number of roads, and the number of special fields.
The second line contains k integers a_1, a_2, …, a_k (1 ≤ a_i ≤ n) — the special fields. All a_i are distinct.
The i-th of the following m lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), representing a bidirectional road between fields x_i and y_i.
It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.
Output
Output one integer, the maximum possible length of the shortest path from field 1 to n after Farmer John installs one road optimally.
Examples
Input
5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
Output
3
Input
5 4 2
2 4
1 2
2 3
3 4
4 5
Output
3
Note
The graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields 3 and 5, and the resulting shortest path from 1 to 5 is length 3.
<image>
The graph for the second example is shown below. Farmer John must add a road between fields 2 and 4, and the resulting shortest path from 1 to 5 is length 3.
<image>
Submitted Solution:
```
import sys
input=sys.stdin.readline
import collections
from collections import defaultdict
def root(k):
count=0
shortest=[]
if l1[k]==1:
count+=1
shortest.append(outta[k])
outta[k]=0
while (k)!=kutta[k]:
k=kutta[k]
if l1[k]==1:
count+=1
shortest.append(outta[k])
outta[k]=0
#print(count,shortest,outta)
return(count,shortest)
def bfs(graph, root):
visited=set()
spec=[0]*(n+1)
parent={root:root}
if l1[root]==1:
spec[1]=0
dist=[0]*(n+1)
count=0
queue=collections.deque([root])
visited.add(root)
while queue:
vertex=queue.popleft()
for neighbour in graph[vertex]:
if neighbour not in visited:
parent[neighbour]=vertex
if l1[neighbour]==1:
spec[neighbour]=dist[vertex]+1
dist[neighbour]=dist[vertex]+1
visited.add(neighbour)
queue.append(neighbour)
return(spec,dist,parent)
n,m,k=map(int,input().split())
special=[int(i) for i in input().split()]
l1=[0]*(n+1)
for j in special:
l1[j]+=1
graph=defaultdict(list)
ok=False
for i in range(m):
x,y=map(int,input().split())
if (x==1 and y==n) or (y==1 and x==n):
ok=True
graph[x].append(y)
graph[y].append(x)
outta,putta,kutta=(bfs(graph,1))
copy=[]
if l1[1]==1:
copy=[0]
for jk in outta:
if jk>0:
copy.append(jk)
xy,yx=root(n)
#print(outta,putta,kutta)
if ok==True or ((k>xy)and max(outta)>=min(yx)):
#print(outta,yx)
print(putta[n])
else:
min_=1000000
copy.sort()
#print(copy,outta,yx)
for j in range(1,len(copy)):
min_=min(min_,copy[j]-copy[j-1])
print(putta[n]-(min_)+1)
``` | instruction | 0 | 53,216 | 1 | 106,432 |
No | output | 1 | 53,216 | 1 | 106,433 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bessie is out grazing on the farm, which consists of n fields connected by m bidirectional roads. She is currently at field 1, and will return to her home at field n at the end of the day.
The Cowfederation of Barns has ordered Farmer John to install one extra bidirectional road. The farm has k special fields and he has decided to install the road between two different special fields. He may add the road between two special fields that already had a road directly connecting them.
After the road is added, Bessie will return home on the shortest path from field 1 to field n. Since Bessie needs more exercise, Farmer John must maximize the length of this shortest path. Help him!
Input
The first line contains integers n, m, and k (2 ≤ n ≤ 2 ⋅ 10^5, n-1 ≤ m ≤ 2 ⋅ 10^5, 2 ≤ k ≤ n) — the number of fields on the farm, the number of roads, and the number of special fields.
The second line contains k integers a_1, a_2, …, a_k (1 ≤ a_i ≤ n) — the special fields. All a_i are distinct.
The i-th of the following m lines contains integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), representing a bidirectional road between fields x_i and y_i.
It is guaranteed that one can reach any field from every other field. It is also guaranteed that for any pair of fields there is at most one road connecting them.
Output
Output one integer, the maximum possible length of the shortest path from field 1 to n after Farmer John installs one road optimally.
Examples
Input
5 5 3
1 3 5
1 2
2 3
3 4
3 5
2 4
Output
3
Input
5 4 2
2 4
1 2
2 3
3 4
4 5
Output
3
Note
The graph for the first example is shown below. The special fields are denoted by red. It is optimal for Farmer John to add a road between fields 3 and 5, and the resulting shortest path from 1 to 5 is length 3.
<image>
The graph for the second example is shown below. Farmer John must add a road between fields 2 and 4, and the resulting shortest path from 1 to 5 is length 3.
<image>
Submitted Solution:
```
import os
from io import BytesIO, IOBase
import sys
from collections import defaultdict,deque
import math
def main():
n, m, k = map(int, input().split())
a = list(map(int, input().split()))
tree = defaultdict(list)
for i in range(m):
x, y = map(int, input().split())
tree[x].append(y)
tree[y].append(x)
d = [0] * (n + 1)
v = [0] * (n + 1)
q = deque()
q.append((0, 1))
v[1] = 1
while q:
p, c = q.pop()
for i in tree[c]:
if (not v[i]):
v[i] = 1
q.appendleft((c, i))
d[i] = d[c] + 1
b = []
for i in a:
if d[i] != 0 or (d[i] == 0 and i == 1):
b.append(i)
b.sort(key=lambda x: d[x])
mi = math.inf
for i in range(len(b) - 1):
z = d[b[i + 1]] - d[b[i]]
if mi > z:
mi = z
print(d[n] - mi + 1)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | instruction | 0 | 53,217 | 1 | 106,434 |
No | output | 1 | 53,217 | 1 | 106,435 |
Provide a correct Python 3 solution for this coding contest problem.
Alice lives on a line. Today, she will travel to some place in a mysterious vehicle. Initially, the distance between Alice and her destination is D. When she input a number x to the vehicle, it will travel in the direction of the destination by a distance of x if this move would shorten the distance between the vehicle and the destination, and it will stay at its position otherwise. Note that the vehicle may go past the destination when the distance between the vehicle and the destination is less than x.
Alice made a list of N numbers. The i-th number in this list is d_i. She will insert these numbers to the vehicle one by one.
However, a mischievous witch appeared. She is thinking of rewriting one number in the list so that Alice will not reach the destination after N moves.
She has Q plans to do this, as follows:
* Rewrite only the q_i-th number in the list with some integer so that Alice will not reach the destination.
Write a program to determine whether each plan is feasible.
Constraints
* 1≤ N ≤ 5*10^5
* 1≤ Q ≤ 5*10^5
* 1≤ D ≤ 10^9
* 1≤ d_i ≤ 10^9(1≤i≤N)
* 1≤ q_i ≤ N(1≤i≤Q)
* D and each d_i are integers.
Input
Input is given from Standard Input in the following format:
N D
d_1 d_2 ... d_N
Q
q_1 q_2 ... q_Q
Output
Print Q lines. The i-th line should contain `YES` if the i-th plan is feasible, and `NO` otherwise.
Examples
Input
4 10
3 4 3 3
2
4 3
Output
NO
YES
Input
5 9
4 4 2 3 2
5
1 4 2 3 5
Output
YES
YES
YES
YES
YES
Input
6 15
4 3 5 4 2 1
6
1 2 3 4 5 6
Output
NO
NO
YES
NO
NO
YES | instruction | 0 | 53,783 | 1 | 107,566 |
"Correct Solution:
```
n, D = map(int, input().split())
x = list(map(int, input().split()))
Q = int(input())
q = list(map(int, input().split()))
ind = [D]
for i in range(n):
ind.append(min(ind[-1], abs(ind[-1] - x[i])))
#print(ind)
l = [1]
for i in range(n-1, -1, -1):
if x[i] < 2*l[-1]:
l.append(x[i] + l[-1])
else:
l.append(l[-1])
l = l[::-1]
for i in range(Q):
if l[q[i]] <= ind[q[i]-1]:
print("YES")
else:
print("NO")
``` | output | 1 | 53,783 | 1 | 107,567 |
Provide a correct Python 3 solution for this coding contest problem.
Alice lives on a line. Today, she will travel to some place in a mysterious vehicle. Initially, the distance between Alice and her destination is D. When she input a number x to the vehicle, it will travel in the direction of the destination by a distance of x if this move would shorten the distance between the vehicle and the destination, and it will stay at its position otherwise. Note that the vehicle may go past the destination when the distance between the vehicle and the destination is less than x.
Alice made a list of N numbers. The i-th number in this list is d_i. She will insert these numbers to the vehicle one by one.
However, a mischievous witch appeared. She is thinking of rewriting one number in the list so that Alice will not reach the destination after N moves.
She has Q plans to do this, as follows:
* Rewrite only the q_i-th number in the list with some integer so that Alice will not reach the destination.
Write a program to determine whether each plan is feasible.
Constraints
* 1≤ N ≤ 5*10^5
* 1≤ Q ≤ 5*10^5
* 1≤ D ≤ 10^9
* 1≤ d_i ≤ 10^9(1≤i≤N)
* 1≤ q_i ≤ N(1≤i≤Q)
* D and each d_i are integers.
Input
Input is given from Standard Input in the following format:
N D
d_1 d_2 ... d_N
Q
q_1 q_2 ... q_Q
Output
Print Q lines. The i-th line should contain `YES` if the i-th plan is feasible, and `NO` otherwise.
Examples
Input
4 10
3 4 3 3
2
4 3
Output
NO
YES
Input
5 9
4 4 2 3 2
5
1 4 2 3 5
Output
YES
YES
YES
YES
YES
Input
6 15
4 3 5 4 2 1
6
1 2 3 4 5 6
Output
NO
NO
YES
NO
NO
YES | instruction | 0 | 53,784 | 1 | 107,568 |
"Correct Solution:
```
N,D=map(int,input().split())
d=list(map(int,input().split()))
dp=[D]*N
if abs(D-d[0])<D:
dp[0]=abs(D-d[0])
for i in range(1,N):
dp[i]=min(dp[i-1],abs(dp[i-1]-d[i]))
ans=["NO"]*N
data=[0]*(N+1)
for i in range(N-1,0,-1):
if d[i]//2>data[i+1]:
data[i]=data[i+1]
else:
data[i]=data[i+1]+d[i]
if dp[i-1]>data[i+1]:
ans[i]="YES"
i=0
if d[i]//2>data[i+1]:
data[i]=data[i+1]
else:
data[i]=data[i+1]+d[i]
if D>data[i+1]:
ans[i]="YES"
Q=int(input())
q=list(map(int,input().split()))
for i in range(Q):
print(ans[q[i]-1])
#print(dp)
#print(data)
#print(ans)
``` | output | 1 | 53,784 | 1 | 107,569 |
Provide a correct Python 3 solution for this coding contest problem.
Alice lives on a line. Today, she will travel to some place in a mysterious vehicle. Initially, the distance between Alice and her destination is D. When she input a number x to the vehicle, it will travel in the direction of the destination by a distance of x if this move would shorten the distance between the vehicle and the destination, and it will stay at its position otherwise. Note that the vehicle may go past the destination when the distance between the vehicle and the destination is less than x.
Alice made a list of N numbers. The i-th number in this list is d_i. She will insert these numbers to the vehicle one by one.
However, a mischievous witch appeared. She is thinking of rewriting one number in the list so that Alice will not reach the destination after N moves.
She has Q plans to do this, as follows:
* Rewrite only the q_i-th number in the list with some integer so that Alice will not reach the destination.
Write a program to determine whether each plan is feasible.
Constraints
* 1≤ N ≤ 5*10^5
* 1≤ Q ≤ 5*10^5
* 1≤ D ≤ 10^9
* 1≤ d_i ≤ 10^9(1≤i≤N)
* 1≤ q_i ≤ N(1≤i≤Q)
* D and each d_i are integers.
Input
Input is given from Standard Input in the following format:
N D
d_1 d_2 ... d_N
Q
q_1 q_2 ... q_Q
Output
Print Q lines. The i-th line should contain `YES` if the i-th plan is feasible, and `NO` otherwise.
Examples
Input
4 10
3 4 3 3
2
4 3
Output
NO
YES
Input
5 9
4 4 2 3 2
5
1 4 2 3 5
Output
YES
YES
YES
YES
YES
Input
6 15
4 3 5 4 2 1
6
1 2 3 4 5 6
Output
NO
NO
YES
NO
NO
YES | instruction | 0 | 53,785 | 1 | 107,570 |
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
N,X = map(int,readline().split())
D = [0] + list(map(int,readline().split()))
readline()
Q = list(map(int,read().split()))
# 各日の出発座標
start = [X] * (N+2)
for n,d in enumerate(D[1:],1):
x = start[n]
y = min(x, abs(x - d))
start[n+1] = y
# 各日において、失敗に至る出発座標の最小値
min_ng = [1] * (N+2)
for n in range(N,0,-1):
d = D[n]
x = min_ng[n+1]
y = x + d
if x + x <= d:
y = x
min_ng[n] = y
answer = ['YES' if x >= y else 'NO' for x,y in zip(start, min_ng[1:])]
print('\n'.join(answer[q] for q in Q))
``` | output | 1 | 53,785 | 1 | 107,571 |
Provide a correct Python 3 solution for this coding contest problem.
Alice lives on a line. Today, she will travel to some place in a mysterious vehicle. Initially, the distance between Alice and her destination is D. When she input a number x to the vehicle, it will travel in the direction of the destination by a distance of x if this move would shorten the distance between the vehicle and the destination, and it will stay at its position otherwise. Note that the vehicle may go past the destination when the distance between the vehicle and the destination is less than x.
Alice made a list of N numbers. The i-th number in this list is d_i. She will insert these numbers to the vehicle one by one.
However, a mischievous witch appeared. She is thinking of rewriting one number in the list so that Alice will not reach the destination after N moves.
She has Q plans to do this, as follows:
* Rewrite only the q_i-th number in the list with some integer so that Alice will not reach the destination.
Write a program to determine whether each plan is feasible.
Constraints
* 1≤ N ≤ 5*10^5
* 1≤ Q ≤ 5*10^5
* 1≤ D ≤ 10^9
* 1≤ d_i ≤ 10^9(1≤i≤N)
* 1≤ q_i ≤ N(1≤i≤Q)
* D and each d_i are integers.
Input
Input is given from Standard Input in the following format:
N D
d_1 d_2 ... d_N
Q
q_1 q_2 ... q_Q
Output
Print Q lines. The i-th line should contain `YES` if the i-th plan is feasible, and `NO` otherwise.
Examples
Input
4 10
3 4 3 3
2
4 3
Output
NO
YES
Input
5 9
4 4 2 3 2
5
1 4 2 3 5
Output
YES
YES
YES
YES
YES
Input
6 15
4 3 5 4 2 1
6
1 2 3 4 5 6
Output
NO
NO
YES
NO
NO
YES | instruction | 0 | 53,786 | 1 | 107,572 |
"Correct Solution:
```
N, D = map( int, input().split() )
d = list( map( int, input().split() ) )
Q = int( input() )
q = list( map( int, input().split() ) )
dis = [ 0 for i in range( N + 1 ) ]
dis[ 0 ] = D
for i in range( N ):
dis[ i + 1 ] = min( dis[ i ], abs( dis[ i ] - d[ i ] ) )
dp = [ 0 for i in range( N + 1 ) ]
dp[ N ] = 1
for i in range( N - 1, -1, -1 ):
if dp[ i + 1 ] <= d[ i ] // 2:
dp[ i ] = dp[ i + 1 ]
else:
dp[ i ] = dp[ i + 1 ] + d[ i ]
for qi in range( Q ):
print( [ "NO", "YES" ][ dis[ q[ qi ] - 1 ] >= dp[ q[ qi ] ] ] )
``` | output | 1 | 53,786 | 1 | 107,573 |
Provide a correct Python 3 solution for this coding contest problem.
Alice lives on a line. Today, she will travel to some place in a mysterious vehicle. Initially, the distance between Alice and her destination is D. When she input a number x to the vehicle, it will travel in the direction of the destination by a distance of x if this move would shorten the distance between the vehicle and the destination, and it will stay at its position otherwise. Note that the vehicle may go past the destination when the distance between the vehicle and the destination is less than x.
Alice made a list of N numbers. The i-th number in this list is d_i. She will insert these numbers to the vehicle one by one.
However, a mischievous witch appeared. She is thinking of rewriting one number in the list so that Alice will not reach the destination after N moves.
She has Q plans to do this, as follows:
* Rewrite only the q_i-th number in the list with some integer so that Alice will not reach the destination.
Write a program to determine whether each plan is feasible.
Constraints
* 1≤ N ≤ 5*10^5
* 1≤ Q ≤ 5*10^5
* 1≤ D ≤ 10^9
* 1≤ d_i ≤ 10^9(1≤i≤N)
* 1≤ q_i ≤ N(1≤i≤Q)
* D and each d_i are integers.
Input
Input is given from Standard Input in the following format:
N D
d_1 d_2 ... d_N
Q
q_1 q_2 ... q_Q
Output
Print Q lines. The i-th line should contain `YES` if the i-th plan is feasible, and `NO` otherwise.
Examples
Input
4 10
3 4 3 3
2
4 3
Output
NO
YES
Input
5 9
4 4 2 3 2
5
1 4 2 3 5
Output
YES
YES
YES
YES
YES
Input
6 15
4 3 5 4 2 1
6
1 2 3 4 5 6
Output
NO
NO
YES
NO
NO
YES | instruction | 0 | 53,787 | 1 | 107,574 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def main():
# YouTubeの通り
dn, s = MI()
dd = LI()
qn = II()
qq = LI()
qi = [(q - 1, i) for i, q in enumerate(qq)]
# pos[i]...dd[i]の移動前の位置
pos = [s]
for d in dd:
pos.append(min(pos[-1], abs(d - pos[-1])))
# print(pos)
# bb[i]...dd[i]の移動前にこの距離以上の位置にいればゴールに届かないという境目
bb = [1] * (dn + 1)
for i in range(dn - 1, -1, -1):
d = dd[i]
if d < bb[i + 1] * 2:
bb[i] = bb[i + 1] + d
else:
bb[i] = bb[i + 1]
# print(bb)
# dd0 dd1 dd2 dd3
# po0 po1 po2 po3 po4
# bb0 bb1 bb2 bb3 bb4
# という前後関係なのでq番目(0-origin)のddに魔法をかけるときは
# pos[q]>=bb[q+1]であるなら妨害可能(YES)
ans = [""] * qn
for q, ai in qi:
if pos[q] >= bb[q + 1]: ans[ai] = "YES"
else: ans[ai] = "NO"
print(*ans, sep="\n")
main()
``` | output | 1 | 53,787 | 1 | 107,575 |
Provide a correct Python 3 solution for this coding contest problem.
Alice lives on a line. Today, she will travel to some place in a mysterious vehicle. Initially, the distance between Alice and her destination is D. When she input a number x to the vehicle, it will travel in the direction of the destination by a distance of x if this move would shorten the distance between the vehicle and the destination, and it will stay at its position otherwise. Note that the vehicle may go past the destination when the distance between the vehicle and the destination is less than x.
Alice made a list of N numbers. The i-th number in this list is d_i. She will insert these numbers to the vehicle one by one.
However, a mischievous witch appeared. She is thinking of rewriting one number in the list so that Alice will not reach the destination after N moves.
She has Q plans to do this, as follows:
* Rewrite only the q_i-th number in the list with some integer so that Alice will not reach the destination.
Write a program to determine whether each plan is feasible.
Constraints
* 1≤ N ≤ 5*10^5
* 1≤ Q ≤ 5*10^5
* 1≤ D ≤ 10^9
* 1≤ d_i ≤ 10^9(1≤i≤N)
* 1≤ q_i ≤ N(1≤i≤Q)
* D and each d_i are integers.
Input
Input is given from Standard Input in the following format:
N D
d_1 d_2 ... d_N
Q
q_1 q_2 ... q_Q
Output
Print Q lines. The i-th line should contain `YES` if the i-th plan is feasible, and `NO` otherwise.
Examples
Input
4 10
3 4 3 3
2
4 3
Output
NO
YES
Input
5 9
4 4 2 3 2
5
1 4 2 3 5
Output
YES
YES
YES
YES
YES
Input
6 15
4 3 5 4 2 1
6
1 2 3 4 5 6
Output
NO
NO
YES
NO
NO
YES | instruction | 0 | 53,788 | 1 | 107,576 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
"""
後ろから、到達不可能距離の集合を見ていきたい。
集合を持つと厳しいが、最小値だけ持っておけばよい。
"""
N,dist = map(int,input().split())
D = [int(x) for x in input().split()]
# 各ターンの出発位置
start_dist = [dist]
for d in D:
x = start_dist[-1]
y = min(abs(x - d), x)
start_dist.append(y)
ng_dist = [None] * (N+1)
ng_dist[N] = 1
for i,d in enumerate(D[::-1]):
x = ng_dist[N-i]
y = x if x <= d//2 else x + d
ng_dist[N-i-1] = y
input()
Q = [int(x) for x in input().split()]
answer = ['YES' if ng_dist[d] <= start_dist[d-1] else 'NO' for d in Q]
print('\n'.join(answer))
``` | output | 1 | 53,788 | 1 | 107,577 |
Provide a correct Python 3 solution for this coding contest problem.
Alice lives on a line. Today, she will travel to some place in a mysterious vehicle. Initially, the distance between Alice and her destination is D. When she input a number x to the vehicle, it will travel in the direction of the destination by a distance of x if this move would shorten the distance between the vehicle and the destination, and it will stay at its position otherwise. Note that the vehicle may go past the destination when the distance between the vehicle and the destination is less than x.
Alice made a list of N numbers. The i-th number in this list is d_i. She will insert these numbers to the vehicle one by one.
However, a mischievous witch appeared. She is thinking of rewriting one number in the list so that Alice will not reach the destination after N moves.
She has Q plans to do this, as follows:
* Rewrite only the q_i-th number in the list with some integer so that Alice will not reach the destination.
Write a program to determine whether each plan is feasible.
Constraints
* 1≤ N ≤ 5*10^5
* 1≤ Q ≤ 5*10^5
* 1≤ D ≤ 10^9
* 1≤ d_i ≤ 10^9(1≤i≤N)
* 1≤ q_i ≤ N(1≤i≤Q)
* D and each d_i are integers.
Input
Input is given from Standard Input in the following format:
N D
d_1 d_2 ... d_N
Q
q_1 q_2 ... q_Q
Output
Print Q lines. The i-th line should contain `YES` if the i-th plan is feasible, and `NO` otherwise.
Examples
Input
4 10
3 4 3 3
2
4 3
Output
NO
YES
Input
5 9
4 4 2 3 2
5
1 4 2 3 5
Output
YES
YES
YES
YES
YES
Input
6 15
4 3 5 4 2 1
6
1 2 3 4 5 6
Output
NO
NO
YES
NO
NO
YES | instruction | 0 | 53,789 | 1 | 107,578 |
"Correct Solution:
```
N, D = map(int, input().split())
ds = list(map(int, input().split()))
Q = int(input())
querys = list(map(lambda x: int(x) - 1, input().split()))
# ps[i]: i回目の移動後の位置（目的地までの距離）
ps = [0] * (N + 1)
ps[0] = D
for i in range(N):
ps[i + 1] = min(abs(ps[i] - ds[i]), ps[i])
if ps[N] != 0:
print('\n'.join(['YES'] * Q))
exit()
# ms[i]: i番目の移動後にm以下の位置ならば全て、目的地に到達可能であるようなmの最大値
ms = [0] * (N + 1)
for i in range(N)[::-1]:
if ms[i + 1] + 1 >= ds[i] - ms[i + 1]:
ms[i] = ms[i + 1] + ds[i]
else:
ms[i] = ms[i + 1]
for q in querys:
if ps[q] <= ms[q + 1]: # 妨害は不可能
print('NO')
else:
print('YES')
``` | output | 1 | 53,789 | 1 | 107,579 |
Provide a correct Python 3 solution for this coding contest problem.
Alice lives on a line. Today, she will travel to some place in a mysterious vehicle. Initially, the distance between Alice and her destination is D. When she input a number x to the vehicle, it will travel in the direction of the destination by a distance of x if this move would shorten the distance between the vehicle and the destination, and it will stay at its position otherwise. Note that the vehicle may go past the destination when the distance between the vehicle and the destination is less than x.
Alice made a list of N numbers. The i-th number in this list is d_i. She will insert these numbers to the vehicle one by one.
However, a mischievous witch appeared. She is thinking of rewriting one number in the list so that Alice will not reach the destination after N moves.
She has Q plans to do this, as follows:
* Rewrite only the q_i-th number in the list with some integer so that Alice will not reach the destination.
Write a program to determine whether each plan is feasible.
Constraints
* 1≤ N ≤ 5*10^5
* 1≤ Q ≤ 5*10^5
* 1≤ D ≤ 10^9
* 1≤ d_i ≤ 10^9(1≤i≤N)
* 1≤ q_i ≤ N(1≤i≤Q)
* D and each d_i are integers.
Input
Input is given from Standard Input in the following format:
N D
d_1 d_2 ... d_N
Q
q_1 q_2 ... q_Q
Output
Print Q lines. The i-th line should contain `YES` if the i-th plan is feasible, and `NO` otherwise.
Examples
Input
4 10
3 4 3 3
2
4 3
Output
NO
YES
Input
5 9
4 4 2 3 2
5
1 4 2 3 5
Output
YES
YES
YES
YES
YES
Input
6 15
4 3 5 4 2 1
6
1 2 3 4 5 6
Output
NO
NO
YES
NO
NO
YES | instruction | 0 | 53,790 | 1 | 107,580 |
"Correct Solution:
```
N,start = map(int,input().split())
D = list(map(int,input().split()))
input()
Q = list(map(int,input().split()))
Q = [(k-1,i) for i,k in enumerate(Q)]
Q.sort()
P = [start]
for d in D:
a = P[-1]
b = abs(a-d)
P.append(min(a,b))
result = [None]*len(Q)
x = 1
for i,d in zip(reversed(range(len(D))), reversed(D)):
if Q[-1][0] == i:
result[Q[-1][1]] = P[i] >= x
Q.pop()
if not Q:
break
if abs(x-d) < x:
x += d
for r in result:
print('YES' if r else 'NO')
``` | output | 1 | 53,790 | 1 | 107,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice lives on a line. Today, she will travel to some place in a mysterious vehicle. Initially, the distance between Alice and her destination is D. When she input a number x to the vehicle, it will travel in the direction of the destination by a distance of x if this move would shorten the distance between the vehicle and the destination, and it will stay at its position otherwise. Note that the vehicle may go past the destination when the distance between the vehicle and the destination is less than x.
Alice made a list of N numbers. The i-th number in this list is d_i. She will insert these numbers to the vehicle one by one.
However, a mischievous witch appeared. She is thinking of rewriting one number in the list so that Alice will not reach the destination after N moves.
She has Q plans to do this, as follows:
* Rewrite only the q_i-th number in the list with some integer so that Alice will not reach the destination.
Write a program to determine whether each plan is feasible.
Constraints
* 1≤ N ≤ 5*10^5
* 1≤ Q ≤ 5*10^5
* 1≤ D ≤ 10^9
* 1≤ d_i ≤ 10^9(1≤i≤N)
* 1≤ q_i ≤ N(1≤i≤Q)
* D and each d_i are integers.
Input
Input is given from Standard Input in the following format:
N D
d_1 d_2 ... d_N
Q
q_1 q_2 ... q_Q
Output
Print Q lines. The i-th line should contain `YES` if the i-th plan is feasible, and `NO` otherwise.
Examples
Input
4 10
3 4 3 3
2
4 3
Output
NO
YES
Input
5 9
4 4 2 3 2
5
1 4 2 3 5
Output
YES
YES
YES
YES
YES
Input
6 15
4 3 5 4 2 1
6
1 2 3 4 5 6
Output
NO
NO
YES
NO
NO
YES
Submitted Solution:
```
N, D = map(int, input().split())
d = list(map(int, input().split()))
Q = int(input())
q = list(map(lambda x : int(x)-1 , input().split()))
dis = [0 for i in range(N+1)]
dis[0] = D
for i in range(N):
dis[i+1] = min(dis[i], abs(dis[i] - d[i]))
dp = [0 for i in range(N+1)]
dp[N] = 1
for i in range(N-1, -1, -1):
if d[i] // 2 >= dp[i+1]:
dp[i] = dp[i+1]
else:
dp[i] = dp[i+1] + d[i]
for qi in range(Q):
print(["YES", "NO"][dis[q[ qi ]] < dp[q[ qi ] + 1]])
``` | instruction | 0 | 53,791 | 1 | 107,582 |
Yes | output | 1 | 53,791 | 1 | 107,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice lives on a line. Today, she will travel to some place in a mysterious vehicle. Initially, the distance between Alice and her destination is D. When she input a number x to the vehicle, it will travel in the direction of the destination by a distance of x if this move would shorten the distance between the vehicle and the destination, and it will stay at its position otherwise. Note that the vehicle may go past the destination when the distance between the vehicle and the destination is less than x.
Alice made a list of N numbers. The i-th number in this list is d_i. She will insert these numbers to the vehicle one by one.
However, a mischievous witch appeared. She is thinking of rewriting one number in the list so that Alice will not reach the destination after N moves.
She has Q plans to do this, as follows:
* Rewrite only the q_i-th number in the list with some integer so that Alice will not reach the destination.
Write a program to determine whether each plan is feasible.
Constraints
* 1≤ N ≤ 5*10^5
* 1≤ Q ≤ 5*10^5
* 1≤ D ≤ 10^9
* 1≤ d_i ≤ 10^9(1≤i≤N)
* 1≤ q_i ≤ N(1≤i≤Q)
* D and each d_i are integers.
Input
Input is given from Standard Input in the following format:
N D
d_1 d_2 ... d_N
Q
q_1 q_2 ... q_Q
Output
Print Q lines. The i-th line should contain `YES` if the i-th plan is feasible, and `NO` otherwise.
Examples
Input
4 10
3 4 3 3
2
4 3
Output
NO
YES
Input
5 9
4 4 2 3 2
5
1 4 2 3 5
Output
YES
YES
YES
YES
YES
Input
6 15
4 3 5 4 2 1
6
1 2 3 4 5 6
Output
NO
NO
YES
NO
NO
YES
Submitted Solution:
```
N, D = list(map(int,input().split()))
d = list(map(int,input().split()))
Q = int(input())
q = list(map(int,input().split()))
a = [D for i in range(N)]
b = [1 for i in range(N+1)]
for i in range(1,N):
a[i] = min(abs(a[i-1]-d[i-1]),a[i-1])
for i in range(N)[::-1]:
if d[i]//2 < b[i+1]:
b[i] = b[i+1] + d[i]
else:
b[i] = b[i+1]
res = ""
for i in q:
if a[i-1] < b[i]:
res+="NO"
else:
res+="YES"
res+="\n"
print(res)
``` | instruction | 0 | 53,792 | 1 | 107,584 |
Yes | output | 1 | 53,792 | 1 | 107,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice lives on a line. Today, she will travel to some place in a mysterious vehicle. Initially, the distance between Alice and her destination is D. When she input a number x to the vehicle, it will travel in the direction of the destination by a distance of x if this move would shorten the distance between the vehicle and the destination, and it will stay at its position otherwise. Note that the vehicle may go past the destination when the distance between the vehicle and the destination is less than x.
Alice made a list of N numbers. The i-th number in this list is d_i. She will insert these numbers to the vehicle one by one.
However, a mischievous witch appeared. She is thinking of rewriting one number in the list so that Alice will not reach the destination after N moves.
She has Q plans to do this, as follows:
* Rewrite only the q_i-th number in the list with some integer so that Alice will not reach the destination.
Write a program to determine whether each plan is feasible.
Constraints
* 1≤ N ≤ 5*10^5
* 1≤ Q ≤ 5*10^5
* 1≤ D ≤ 10^9
* 1≤ d_i ≤ 10^9(1≤i≤N)
* 1≤ q_i ≤ N(1≤i≤Q)
* D and each d_i are integers.
Input
Input is given from Standard Input in the following format:
N D
d_1 d_2 ... d_N
Q
q_1 q_2 ... q_Q
Output
Print Q lines. The i-th line should contain `YES` if the i-th plan is feasible, and `NO` otherwise.
Examples
Input
4 10
3 4 3 3
2
4 3
Output
NO
YES
Input
5 9
4 4 2 3 2
5
1 4 2 3 5
Output
YES
YES
YES
YES
YES
Input
6 15
4 3 5 4 2 1
6
1 2 3 4 5 6
Output
NO
NO
YES
NO
NO
YES
Submitted Solution:
```
n, D = map(int, input().split())
x = list(map(int, input().split()))
Q = int(input())
q = list(map(int, input().split()))
ind = [D]
for i in range(n):
ind.append(min(ind[-1], abs(ind[-1] - x[i])))
l = [1]
for i in range(n-1, -1, -1):
if x[i] < 2*l[-1]:
l.append(x[i] + l[-1])
else:
l.append(l[-1])
l = l[::-1]
for i in range(Q):
if l[q[i]] <= ind[q[i]-1]:
print("YES")
else:
print("NO")
``` | instruction | 0 | 53,793 | 1 | 107,586 |
Yes | output | 1 | 53,793 | 1 | 107,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice lives on a line. Today, she will travel to some place in a mysterious vehicle. Initially, the distance between Alice and her destination is D. When she input a number x to the vehicle, it will travel in the direction of the destination by a distance of x if this move would shorten the distance between the vehicle and the destination, and it will stay at its position otherwise. Note that the vehicle may go past the destination when the distance between the vehicle and the destination is less than x.
Alice made a list of N numbers. The i-th number in this list is d_i. She will insert these numbers to the vehicle one by one.
However, a mischievous witch appeared. She is thinking of rewriting one number in the list so that Alice will not reach the destination after N moves.
She has Q plans to do this, as follows:
* Rewrite only the q_i-th number in the list with some integer so that Alice will not reach the destination.
Write a program to determine whether each plan is feasible.
Constraints
* 1≤ N ≤ 5*10^5
* 1≤ Q ≤ 5*10^5
* 1≤ D ≤ 10^9
* 1≤ d_i ≤ 10^9(1≤i≤N)
* 1≤ q_i ≤ N(1≤i≤Q)
* D and each d_i are integers.
Input
Input is given from Standard Input in the following format:
N D
d_1 d_2 ... d_N
Q
q_1 q_2 ... q_Q
Output
Print Q lines. The i-th line should contain `YES` if the i-th plan is feasible, and `NO` otherwise.
Examples
Input
4 10
3 4 3 3
2
4 3
Output
NO
YES
Input
5 9
4 4 2 3 2
5
1 4 2 3 5
Output
YES
YES
YES
YES
YES
Input
6 15
4 3 5 4 2 1
6
1 2 3 4 5 6
Output
NO
NO
YES
NO
NO
YES
Submitted Solution:
```
# E
N, D = map(int, input().split())
dv = list(map(int, input().split()))
Q = int(input())
qv = list(map(int, input().split()))
# trace of Alice
dist_alice = D
dist_alice_list = [D]
for i in range(N):
if dist_alice >= dv[i]:
dist_alice -= dv[i]
elif 2*dist_alice >= dv[i]:
dist_alice = dv[i] - dist_alice
dist_alice_list.append(dist_alice)
sol_min = [1]
sol_min_ = 1
for i in range(N-1, -1, -1):
if 2 * sol_min_ > dv[i]:
sol_min_ += dv[i]
sol_min.append(sol_min_)
for i in range(Q):
if dist_alice_list[qv[i]-1] >= sol_min[N-qv[i]]:
print("YES")
else:
print("NO")
``` | instruction | 0 | 53,794 | 1 | 107,588 |
Yes | output | 1 | 53,794 | 1 | 107,589 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice lives on a line. Today, she will travel to some place in a mysterious vehicle. Initially, the distance between Alice and her destination is D. When she input a number x to the vehicle, it will travel in the direction of the destination by a distance of x if this move would shorten the distance between the vehicle and the destination, and it will stay at its position otherwise. Note that the vehicle may go past the destination when the distance between the vehicle and the destination is less than x.
Alice made a list of N numbers. The i-th number in this list is d_i. She will insert these numbers to the vehicle one by one.
However, a mischievous witch appeared. She is thinking of rewriting one number in the list so that Alice will not reach the destination after N moves.
She has Q plans to do this, as follows:
* Rewrite only the q_i-th number in the list with some integer so that Alice will not reach the destination.
Write a program to determine whether each plan is feasible.
Constraints
* 1≤ N ≤ 5*10^5
* 1≤ Q ≤ 5*10^5
* 1≤ D ≤ 10^9
* 1≤ d_i ≤ 10^9(1≤i≤N)
* 1≤ q_i ≤ N(1≤i≤Q)
* D and each d_i are integers.
Input
Input is given from Standard Input in the following format:
N D
d_1 d_2 ... d_N
Q
q_1 q_2 ... q_Q
Output
Print Q lines. The i-th line should contain `YES` if the i-th plan is feasible, and `NO` otherwise.
Examples
Input
4 10
3 4 3 3
2
4 3
Output
NO
YES
Input
5 9
4 4 2 3 2
5
1 4 2 3 5
Output
YES
YES
YES
YES
YES
Input
6 15
4 3 5 4 2 1
6
1 2 3 4 5 6
Output
NO
NO
YES
NO
NO
YES
Submitted Solution:
```
n, d = [int(x) for x in input().split(" ")]
arr = [int(x) for x in input().split(" ")]
q = int(input())
queries = [int(x) for x in input().split(" ")]
A = [0]*n
currDist = d
before = True
done = False
for i in range(len(arr)):
if done:
A[i] = 0
elif arr[i] < currDist*2:
if arr[i] == currDist:
done = True
A[i] = 0
currDist = 0
elif arr[i] < currDist:
A[i] = currDist - arr[i]
currDist = A[i]
else:
A[i] = currDist
B = [0]*n
B[n-1] = 1
for i in range(n-1):
j = n-2-i
if arr[j] >= 2*B[j+1]:
B[j] = B[j+1]
else:
B[j] = B[j+1] + arr[j]
for i in range(q):
pos = queries[i] - 1
if pos == 0:
print("YES" if (B[0] <= d) else "NO")
else:
print("YES" if (B[pos] <= A[pos-1]) else "NO")
``` | instruction | 0 | 53,795 | 1 | 107,590 |
No | output | 1 | 53,795 | 1 | 107,591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice lives on a line. Today, she will travel to some place in a mysterious vehicle. Initially, the distance between Alice and her destination is D. When she input a number x to the vehicle, it will travel in the direction of the destination by a distance of x if this move would shorten the distance between the vehicle and the destination, and it will stay at its position otherwise. Note that the vehicle may go past the destination when the distance between the vehicle and the destination is less than x.
Alice made a list of N numbers. The i-th number in this list is d_i. She will insert these numbers to the vehicle one by one.
However, a mischievous witch appeared. She is thinking of rewriting one number in the list so that Alice will not reach the destination after N moves.
She has Q plans to do this, as follows:
* Rewrite only the q_i-th number in the list with some integer so that Alice will not reach the destination.
Write a program to determine whether each plan is feasible.
Constraints
* 1≤ N ≤ 5*10^5
* 1≤ Q ≤ 5*10^5
* 1≤ D ≤ 10^9
* 1≤ d_i ≤ 10^9(1≤i≤N)
* 1≤ q_i ≤ N(1≤i≤Q)
* D and each d_i are integers.
Input
Input is given from Standard Input in the following format:
N D
d_1 d_2 ... d_N
Q
q_1 q_2 ... q_Q
Output
Print Q lines. The i-th line should contain `YES` if the i-th plan is feasible, and `NO` otherwise.
Examples
Input
4 10
3 4 3 3
2
4 3
Output
NO
YES
Input
5 9
4 4 2 3 2
5
1 4 2 3 5
Output
YES
YES
YES
YES
YES
Input
6 15
4 3 5 4 2 1
6
1 2 3 4 5 6
Output
NO
NO
YES
NO
NO
YES
Submitted Solution:
```
import sys
input = sys.stdin.readline
"""
後ろから、到達不可能距離の集合を見ていきたい。
集合を持つと厳しいが、最小値だけ持っておけばよい。
"""
N,dist = map(int,input().split())
D = [int(x) for x in input().split()]
# 各ターンの出発位置
start_dist = [dist]
for d in D:
x = start_dist[-1]
y = min(abs(x - d), x)
start_dist.append(y)
ng_dist = [None] * (N+1)
ng_dist[N] = 1
for i,d in enumerate(D[::-1]):
x = ng_dist[N-i]
y = x if x < d else x + d
ng_dist[N-i-1] = y
input()
Q = [int(x) for x in input().split()]
answer = ['YES' if ng_dist[d] <= start_dist[d-1] else 'NO' for d in Q]
print('\n'.join(answer))
``` | instruction | 0 | 53,796 | 1 | 107,592 |
No | output | 1 | 53,796 | 1 | 107,593 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice lives on a line. Today, she will travel to some place in a mysterious vehicle. Initially, the distance between Alice and her destination is D. When she input a number x to the vehicle, it will travel in the direction of the destination by a distance of x if this move would shorten the distance between the vehicle and the destination, and it will stay at its position otherwise. Note that the vehicle may go past the destination when the distance between the vehicle and the destination is less than x.
Alice made a list of N numbers. The i-th number in this list is d_i. She will insert these numbers to the vehicle one by one.
However, a mischievous witch appeared. She is thinking of rewriting one number in the list so that Alice will not reach the destination after N moves.
She has Q plans to do this, as follows:
* Rewrite only the q_i-th number in the list with some integer so that Alice will not reach the destination.
Write a program to determine whether each plan is feasible.
Constraints
* 1≤ N ≤ 5*10^5
* 1≤ Q ≤ 5*10^5
* 1≤ D ≤ 10^9
* 1≤ d_i ≤ 10^9(1≤i≤N)
* 1≤ q_i ≤ N(1≤i≤Q)
* D and each d_i are integers.
Input
Input is given from Standard Input in the following format:
N D
d_1 d_2 ... d_N
Q
q_1 q_2 ... q_Q
Output
Print Q lines. The i-th line should contain `YES` if the i-th plan is feasible, and `NO` otherwise.
Examples
Input
4 10
3 4 3 3
2
4 3
Output
NO
YES
Input
5 9
4 4 2 3 2
5
1 4 2 3 5
Output
YES
YES
YES
YES
YES
Input
6 15
4 3 5 4 2 1
6
1 2 3 4 5 6
Output
NO
NO
YES
NO
NO
YES
Submitted Solution:
```
N, D = map(int, input().split())
d = list(map(int, input().split()))
Q = int(input())
q = list(map(lambda x : int(x)-1 , input().split()))
dis = [0 for i in range(N+1)]
dis[0] = D
for i in range(N):
dis[i+1] = min(dis[i], abs(dis[i] - d[i]))
dp = [0 for i in range(N+1)]
dp[N] = 1
for i in range(N-1, -1, -1):
if d[i] // 2 >= dp[i+1]:
dp[i] = dp[i+1]
else:
dp[i] = dp[i+1] + d[i]
for qi in range(Q):
print(["YES", "NO"][dis[q[ qi ]] <= dp[q[ qi ] + 1]])
``` | instruction | 0 | 53,797 | 1 | 107,594 |
No | output | 1 | 53,797 | 1 | 107,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice lives on a line. Today, she will travel to some place in a mysterious vehicle. Initially, the distance between Alice and her destination is D. When she input a number x to the vehicle, it will travel in the direction of the destination by a distance of x if this move would shorten the distance between the vehicle and the destination, and it will stay at its position otherwise. Note that the vehicle may go past the destination when the distance between the vehicle and the destination is less than x.
Alice made a list of N numbers. The i-th number in this list is d_i. She will insert these numbers to the vehicle one by one.
However, a mischievous witch appeared. She is thinking of rewriting one number in the list so that Alice will not reach the destination after N moves.
She has Q plans to do this, as follows:
* Rewrite only the q_i-th number in the list with some integer so that Alice will not reach the destination.
Write a program to determine whether each plan is feasible.
Constraints
* 1≤ N ≤ 5*10^5
* 1≤ Q ≤ 5*10^5
* 1≤ D ≤ 10^9
* 1≤ d_i ≤ 10^9(1≤i≤N)
* 1≤ q_i ≤ N(1≤i≤Q)
* D and each d_i are integers.
Input
Input is given from Standard Input in the following format:
N D
d_1 d_2 ... d_N
Q
q_1 q_2 ... q_Q
Output
Print Q lines. The i-th line should contain `YES` if the i-th plan is feasible, and `NO` otherwise.
Examples
Input
4 10
3 4 3 3
2
4 3
Output
NO
YES
Input
5 9
4 4 2 3 2
5
1 4 2 3 5
Output
YES
YES
YES
YES
YES
Input
6 15
4 3 5 4 2 1
6
1 2 3 4 5 6
Output
NO
NO
YES
NO
NO
YES
Submitted Solution:
```
x = 1
for i,d in zip(reversed(range(len(D))), reversed(D)):
if Q[-1][0] == i:
result[Q[-1][1]] = P[i] >= x
Q.pop()
if not Q:
break
if abs(x-d) < x:
x += d
for r in result:
print('YES' if r else 'NO')
``` | instruction | 0 | 53,798 | 1 | 107,596 |
No | output | 1 | 53,798 | 1 | 107,597 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
There are N villages. Each village is numbered from 1 to N. Due to the recent merger boom, several villages have been merged. Two or more merged villages will become one new city, and villages that are not merged with any village will remain villages.
You will be given multiple pieces of information that two villages will be in the same city after the merger. Depending on the combination of the information, three or more villages can become one city.
Given information about villages that will be in the same city after the merger, output the absolute value of the difference between the number of cities and the number of villages after the merger.
Constraints
* 1 ≤ N ≤ 1,000
* 0 ≤ M ≤ 100
* 1 ≤ ai ≤ N
* 1 ≤ bi ≤ N
Input
The input is given in the following format.
N M
a1 b1
a2 b2
...
ai bi
...
aM bM
The first line gives the number N of villages and the number M of information about the merger, separated by blanks.
From the second line to the M + 1 line, two integers ai and bi representing information about the merger are given, separated by blanks. Each information indicates that the ai and bi villages will be the same city after the merger. However, no input is given that satisfies ai = bi.
Output
Output the absolute value of the difference between the number of villages and the number of cities on one line.
Examples
Input
3 1
1 2
Output
0
Input
4 2
1 4
2 3
Output
2
Input
5 0
Output
5
Input
3 3
1 2
2 3
3 1
Output
1
Input
3 2
1 2
2 3
Output
1
Input
5 4
1 2
2 3
3 4
4 5
Output
1
Input
10 5
3 4
1 2
9 6
2 6
2 9
Output
2 | instruction | 0 | 53,869 | 1 | 107,738 |
"Correct Solution:
```
n, m = map(int, input().split())
q = []
for _ in range(m):
a, b = map(int, input().split())
i, j = -1, -1
for k in range(len(q)):
if i == -1:
if a in q[k]:
i = k
if j == -1:
if b in q[k]:
j = k
if i >= 0 and j >= 0:
break
if i >= 0:
if j >= 0:
if i != j:
q[i] |= q[j]
q.pop(j)
else:
q[i].add(b)
elif j >= 0:
q[j].add(a)
else:
q.append(set([a, b]))
print(abs(n - sum([len(i) for i in q]) - len(q)))
``` | output | 1 | 53,869 | 1 | 107,739 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
There are N villages. Each village is numbered from 1 to N. Due to the recent merger boom, several villages have been merged. Two or more merged villages will become one new city, and villages that are not merged with any village will remain villages.
You will be given multiple pieces of information that two villages will be in the same city after the merger. Depending on the combination of the information, three or more villages can become one city.
Given information about villages that will be in the same city after the merger, output the absolute value of the difference between the number of cities and the number of villages after the merger.
Constraints
* 1 ≤ N ≤ 1,000
* 0 ≤ M ≤ 100
* 1 ≤ ai ≤ N
* 1 ≤ bi ≤ N
Input
The input is given in the following format.
N M
a1 b1
a2 b2
...
ai bi
...
aM bM
The first line gives the number N of villages and the number M of information about the merger, separated by blanks.
From the second line to the M + 1 line, two integers ai and bi representing information about the merger are given, separated by blanks. Each information indicates that the ai and bi villages will be the same city after the merger. However, no input is given that satisfies ai = bi.
Output
Output the absolute value of the difference between the number of villages and the number of cities on one line.
Examples
Input
3 1
1 2
Output
0
Input
4 2
1 4
2 3
Output
2
Input
5 0
Output
5
Input
3 3
1 2
2 3
3 1
Output
1
Input
3 2
1 2
2 3
Output
1
Input
5 4
1 2
2 3
3 4
4 5
Output
1
Input
10 5
3 4
1 2
9 6
2 6
2 9
Output
2 | instruction | 0 | 53,871 | 1 | 107,742 |
"Correct Solution:
```
[m,n] = map(int,input().split())
def find(v,cs):
for c in cities:
if v in c:
return (True,c)
return (False,set([v]))
cities = []
for _ in range(n):
[a,b] = map(int,input().split())
(ra,fa) = find(a,cities)
(rb,fb) = find(b,cities)
mg = fa | fb
if ra:
cities.remove(fa)
else:
m = m - 1
if rb:
if fa != fb:
cities.remove(fb)
else:
m = m - 1
cities.append(mg)
print(abs(m-len(cities)))
``` | output | 1 | 53,871 | 1 | 107,743 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
There are N villages. Each village is numbered from 1 to N. Due to the recent merger boom, several villages have been merged. Two or more merged villages will become one new city, and villages that are not merged with any village will remain villages.
You will be given multiple pieces of information that two villages will be in the same city after the merger. Depending on the combination of the information, three or more villages can become one city.
Given information about villages that will be in the same city after the merger, output the absolute value of the difference between the number of cities and the number of villages after the merger.
Constraints
* 1 ≤ N ≤ 1,000
* 0 ≤ M ≤ 100
* 1 ≤ ai ≤ N
* 1 ≤ bi ≤ N
Input
The input is given in the following format.
N M
a1 b1
a2 b2
...
ai bi
...
aM bM
The first line gives the number N of villages and the number M of information about the merger, separated by blanks.
From the second line to the M + 1 line, two integers ai and bi representing information about the merger are given, separated by blanks. Each information indicates that the ai and bi villages will be the same city after the merger. However, no input is given that satisfies ai = bi.
Output
Output the absolute value of the difference between the number of villages and the number of cities on one line.
Examples
Input
3 1
1 2
Output
0
Input
4 2
1 4
2 3
Output
2
Input
5 0
Output
5
Input
3 3
1 2
2 3
3 1
Output
1
Input
3 2
1 2
2 3
Output
1
Input
5 4
1 2
2 3
3 4
4 5
Output
1
Input
10 5
3 4
1 2
9 6
2 6
2 9
Output
2 | instruction | 0 | 53,872 | 1 | 107,744 |
"Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import array
class UnionFind(object):
def __init__(self, number_of_nodes, typecode="L"):
self.typecode = typecode
self.par = array.array(typecode, range(number_of_nodes))
self.rank = array.array(typecode, (0 for i in range(number_of_nodes)))
def root(self, node):
if self.par[node] == node:
return node
else:
r = self.root(self.par[node])
self.par[node] = r
return r
def in_the_same_set(self, node1, node2):
return self.root(node1) == self.root(node2)
def unite(self, node1, node2):
x = self.root(node1)
y = self.root(node2)
if x == y:
pass
elif self.rank[x] < self.rank[y]:
self.par[x] = y
else:
self.par[y] = x
if self.rank[x] == self.rank[y]:
self.rank[x] += 1
def main():
n, m = map(int, input().split())
uf = UnionFind(n, "I")
for _ in range(m):
a, b = map(lambda x: int(x) - 1, input().split())
uf.unite(a, b)
city_representatives = set()
num_cities_and_villages = 0
for i in range(n):
rep = uf.root(i)
if i == rep:
num_cities_and_villages += 1
else:
city_representatives.add(rep)
num_cities = len(city_representatives)
answer = abs(num_cities_and_villages - 2 * num_cities)
print(answer)
if __name__ == '__main__':
main()
``` | output | 1 | 53,872 | 1 | 107,745 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
There are N villages. Each village is numbered from 1 to N. Due to the recent merger boom, several villages have been merged. Two or more merged villages will become one new city, and villages that are not merged with any village will remain villages.
You will be given multiple pieces of information that two villages will be in the same city after the merger. Depending on the combination of the information, three or more villages can become one city.
Given information about villages that will be in the same city after the merger, output the absolute value of the difference between the number of cities and the number of villages after the merger.
Constraints
* 1 ≤ N ≤ 1,000
* 0 ≤ M ≤ 100
* 1 ≤ ai ≤ N
* 1 ≤ bi ≤ N
Input
The input is given in the following format.
N M
a1 b1
a2 b2
...
ai bi
...
aM bM
The first line gives the number N of villages and the number M of information about the merger, separated by blanks.
From the second line to the M + 1 line, two integers ai and bi representing information about the merger are given, separated by blanks. Each information indicates that the ai and bi villages will be the same city after the merger. However, no input is given that satisfies ai = bi.
Output
Output the absolute value of the difference between the number of villages and the number of cities on one line.
Examples
Input
3 1
1 2
Output
0
Input
4 2
1 4
2 3
Output
2
Input
5 0
Output
5
Input
3 3
1 2
2 3
3 1
Output
1
Input
3 2
1 2
2 3
Output
1
Input
5 4
1 2
2 3
3 4
4 5
Output
1
Input
10 5
3 4
1 2
9 6
2 6
2 9
Output
2 | instruction | 0 | 53,873 | 1 | 107,746 |
"Correct Solution:
```
n, m = map(int, input().split())
parent = [i for i in range(n)]
def get_par(x):
if x == parent[x]:return x
parent[x] = get_par(parent[x])
return parent[x]
for _ in range(m):
a, b = map(int, input().split())
a -= 1
b -= 1
pa, pb = get_par(a), get_par(b)
parent[pb] = pa
city_dic = set()
village_cnt = 0
for i in range(n):
pi = get_par(i)
if i == pi:village_cnt += 1
else:city_dic.add(pi)
print(abs(2 * len(city_dic) - village_cnt))
``` | output | 1 | 53,873 | 1 | 107,747 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Farmer John has a farm that consists of n pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.
Unfortunately, Farmer John lost the map of the farm. All he remembers is an array d, where d_i is the smallest amount of time it took the cows to reach the i-th pasture from pasture 1 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of pastures.
The second line of each test case contains n space separated integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the array d. It is guaranteed that d_1 = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory.
Example
Input
3
3
0 2 3
2
0 1000000000
1
0
Output
-3
0
0
Note
In the first test case, you can add roads
* from pasture 1 to pasture 2 with a time of 2,
* from pasture 2 to pasture 3 with a time of 1,
* from pasture 3 to pasture 1 with a time of -3,
* from pasture 3 to pasture 2 with a time of -1,
* from pasture 2 to pasture 1 with a time of -2.
The total cost is 2 + 1 + -3 + -1 + -2 = -3.
In the second test case, you can add a road from pasture 1 to pasture 2 with cost 1000000000 and a road from pasture 2 to pasture 1 with cost -1000000000. The total cost is 1000000000 + -1000000000 = 0.
In the third test case, you can't add any roads. The total cost is 0. | instruction | 0 | 54,200 | 1 | 108,400 |
Tags: constructive algorithms, graphs, greedy, shortest paths, sortings
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
#<fast I/O>
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
#</fast I/O>
#<I/O>
def scanf(datatype = str):
return datatype(sys.stdin.readline().rstrip())
def printf(answer = ''):
return sys.stdout.write(str(answer) + "\n")
def prints(answer):
return sys.stdout.write(str(answer) + " ")
def map_input(datatype = str):
return map(datatype, sys.stdin.readline().split())
def list_input(datatype = str):
return list(map(datatype, sys.stdin.readline().split()))
def testcase(number: int, solve_function):
for _ in range(number):
printf(solve_function())
# solve_function()
#</I/O>
#<solution>
def solve():
n = scanf(int)
arr = list_input(int)
if(n <= 2):
return 0
arr.sort()
prefix = [0] * n
prefix[0] = arr[0]
for i in range(1, n):
prefix[i] = prefix[i - 1] + arr[i]
Sum = 0
mul = 1
for i in range(2, n):
Sum += (arr[i] * mul) - prefix[i - 2]
mul += 1
Sum *= -1
return Sum
# count = 0
# for i in range(1, n + 1):
# for j in range(i + 1, n + 1):
# if(arr[i] * arr[j] == i + j):
# count += 1
# return count
# solve()
t = scanf(int)
testcase(t,solve)
#</solution>
``` | output | 1 | 54,200 | 1 | 108,401 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Farmer John has a farm that consists of n pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.
Unfortunately, Farmer John lost the map of the farm. All he remembers is an array d, where d_i is the smallest amount of time it took the cows to reach the i-th pasture from pasture 1 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of pastures.
The second line of each test case contains n space separated integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the array d. It is guaranteed that d_1 = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory.
Example
Input
3
3
0 2 3
2
0 1000000000
1
0
Output
-3
0
0
Note
In the first test case, you can add roads
* from pasture 1 to pasture 2 with a time of 2,
* from pasture 2 to pasture 3 with a time of 1,
* from pasture 3 to pasture 1 with a time of -3,
* from pasture 3 to pasture 2 with a time of -1,
* from pasture 2 to pasture 1 with a time of -2.
The total cost is 2 + 1 + -3 + -1 + -2 = -3.
In the second test case, you can add a road from pasture 1 to pasture 2 with cost 1000000000 and a road from pasture 2 to pasture 1 with cost -1000000000. The total cost is 1000000000 + -1000000000 = 0.
In the third test case, you can't add any roads. The total cost is 0. | instruction | 0 | 54,201 | 1 | 108,402 |
Tags: constructive algorithms, graphs, greedy, shortest paths, sortings
Correct Solution:
```
"""
ID: happyn61
LANG: PYTHON3
PROB: loan
"""
from itertools import product
import itertools
#from collections import defaultdict
import sys
import heapq
from collections import deque
MOD=1000000000007
#fin = open ('loan.in', 'r')
#fout = open ('loan.out', 'w')
#print(dic["4734"])
def find(parent,i):
if parent[i] != i:
parent[i]=find(parent,parent[i])
return parent[i]
# A utility function to do union of two subsets
def union(parent,rank,xx,yy):
x=find(parent,xx)
y=find(parent,yy)
if rank[x]>rank[y]:
parent[y]=x
elif rank[y]>rank[x]:
parent[x]=y
else:
parent[y]=x
rank[x]+=1
ans=0
#NK=sys.stdin.readline().strip().split()
K=int(sys.stdin.readline().strip())
#N=int(NK[0])
#K=int(NK[1])
#M=int(NK[2])
#ol=list(map(int,sys.stdin.readline().strip().split()))
#d={0:0,1:0}
x=0
y=0
#d={"N":(0,1),"S":(0,-1),"W":(-1,0),"E":(1,0)}
for _ in range(K):
n=int(sys.stdin.readline().strip())
a=list(map(int,sys.stdin.readline().strip().split()))
if n<3:
print("0")
else:
ans=0
a.sort()
p=[0]
for i in range(1,n):
p.append(p[-1]+a[i])
for i in range(2,n):
#print(p,ans,i)
ans+=((i-1)*a[i]-p[i-2])
print(-ans)
``` | output | 1 | 54,201 | 1 | 108,403 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Farmer John has a farm that consists of n pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.
Unfortunately, Farmer John lost the map of the farm. All he remembers is an array d, where d_i is the smallest amount of time it took the cows to reach the i-th pasture from pasture 1 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of pastures.
The second line of each test case contains n space separated integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the array d. It is guaranteed that d_1 = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory.
Example
Input
3
3
0 2 3
2
0 1000000000
1
0
Output
-3
0
0
Note
In the first test case, you can add roads
* from pasture 1 to pasture 2 with a time of 2,
* from pasture 2 to pasture 3 with a time of 1,
* from pasture 3 to pasture 1 with a time of -3,
* from pasture 3 to pasture 2 with a time of -1,
* from pasture 2 to pasture 1 with a time of -2.
The total cost is 2 + 1 + -3 + -1 + -2 = -3.
In the second test case, you can add a road from pasture 1 to pasture 2 with cost 1000000000 and a road from pasture 2 to pasture 1 with cost -1000000000. The total cost is 1000000000 + -1000000000 = 0.
In the third test case, you can't add any roads. The total cost is 0. | instruction | 0 | 54,202 | 1 | 108,404 |
Tags: constructive algorithms, graphs, greedy, shortest paths, sortings
Correct Solution:
```
'''
Rohit Kumar Vishwas
13th June 2021
'''
import sys
import math
import os.path
from collections import *
from queue import *
from bisect import *
from heapq import *
# sys.setrecursionlimit(10 ** 6)
MOD = int(1e9 + 7)
inf = float('inf')
ninf = float('-inf')
input = sys.stdin.readline
def ii(): return list(map(int, input().strip().split()))
def deb(a, k=0, b="Debugging"): print(b, " Test Case no.",
k, " :-", *a) if os.path.exists('inp.txt') else ""
for _ in range(int(input())):
n, = ii()
arr = ii()
arr.sort()
su = arr[-1]
for k in range(1, n):
su -= (arr[k]-arr[k-1])*(n-k)*k
print(su)
``` | output | 1 | 54,202 | 1 | 108,405 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Farmer John has a farm that consists of n pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.
Unfortunately, Farmer John lost the map of the farm. All he remembers is an array d, where d_i is the smallest amount of time it took the cows to reach the i-th pasture from pasture 1 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of pastures.
The second line of each test case contains n space separated integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the array d. It is guaranteed that d_1 = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory.
Example
Input
3
3
0 2 3
2
0 1000000000
1
0
Output
-3
0
0
Note
In the first test case, you can add roads
* from pasture 1 to pasture 2 with a time of 2,
* from pasture 2 to pasture 3 with a time of 1,
* from pasture 3 to pasture 1 with a time of -3,
* from pasture 3 to pasture 2 with a time of -1,
* from pasture 2 to pasture 1 with a time of -2.
The total cost is 2 + 1 + -3 + -1 + -2 = -3.
In the second test case, you can add a road from pasture 1 to pasture 2 with cost 1000000000 and a road from pasture 2 to pasture 1 with cost -1000000000. The total cost is 1000000000 + -1000000000 = 0.
In the third test case, you can't add any roads. The total cost is 0. | instruction | 0 | 54,203 | 1 | 108,406 |
Tags: constructive algorithms, graphs, greedy, shortest paths, sortings
Correct Solution:
```
'''
Auther: ghoshashis545 Ashis Ghosh
College: Jalpaiguri Govt Enggineering College
'''
from os import path
from io import BytesIO, IOBase
import sys
from heapq import heappush,heappop
from functools import cmp_to_key as ctk
from collections import deque,Counter,defaultdict as dd
from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
from itertools import permutations
from datetime import datetime
from math import ceil,sqrt,log,gcd
def ii():return int(input())
def si():return input().rstrip()
def mi():return map(int,input().split())
def li():return list(mi())
abc='abcdefghijklmnopqrstuvwxyz'
mod=1000000007
#mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def bo(i):
return ord(i)-ord('0')
file = 1
def ceil(a,b):
return (a+b-1)//b
def solve():
for _ in range(1,ii()+1):
n = ii()
a = li()
a.sort()
ans = 0
for i in range(1,n):
ans += (a[i]-a[i-1])
s = 0
# print(ans)
for i in range(1,n):
ans += (s-a[i]*i)
s += a[i]
# print(ans)
print(ans)
if __name__ =="__main__":
if(file):
if path.exists('input.txt'):
sys.stdin=open('input.txt', 'r')
sys.stdout=open('output.txt','w')
else:
input=sys.stdin.readline
solve()
``` | output | 1 | 54,203 | 1 | 108,407 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Farmer John has a farm that consists of n pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.
Unfortunately, Farmer John lost the map of the farm. All he remembers is an array d, where d_i is the smallest amount of time it took the cows to reach the i-th pasture from pasture 1 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of pastures.
The second line of each test case contains n space separated integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the array d. It is guaranteed that d_1 = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory.
Example
Input
3
3
0 2 3
2
0 1000000000
1
0
Output
-3
0
0
Note
In the first test case, you can add roads
* from pasture 1 to pasture 2 with a time of 2,
* from pasture 2 to pasture 3 with a time of 1,
* from pasture 3 to pasture 1 with a time of -3,
* from pasture 3 to pasture 2 with a time of -1,
* from pasture 2 to pasture 1 with a time of -2.
The total cost is 2 + 1 + -3 + -1 + -2 = -3.
In the second test case, you can add a road from pasture 1 to pasture 2 with cost 1000000000 and a road from pasture 2 to pasture 1 with cost -1000000000. The total cost is 1000000000 + -1000000000 = 0.
In the third test case, you can't add any roads. The total cost is 0. | instruction | 0 | 54,204 | 1 | 108,408 |
Tags: constructive algorithms, graphs, greedy, shortest paths, sortings
Correct Solution:
```
# Author Name: Ajay Meena
# Codeforce : https://codeforces.com/profile/majay1638
import sys
import math
import bisect
import heapq
from bisect import bisect_right
from sys import stdin, stdout
# -------------- INPUT FUNCTIONS ------------------
def get_ints_in_variables(): return map(
int, sys.stdin.readline().strip().split())
def get_int(): return int(sys.stdin.readline())
def get_ints_in_list(): return list(
map(int, sys.stdin.readline().strip().split()))
def get_list_of_list(n): return [list(
map(int, sys.stdin.readline().strip().split())) for _ in range(n)]
def get_string(): return sys.stdin.readline().strip()
# -------- SOME CUSTOMIZED FUNCTIONS-----------
def myceil(x, y): return (x + y - 1) // y
# -------------- SOLUTION FUNCTION ------------------
def Solution(arr, n):
# Write Your Code Here
arr = sorted(arr)
ans = arr[-1]
prev = 0
for i in range(1, n):
prev = (prev+i*(arr[i]-arr[i-1]))
ans -= prev
print(ans)
def main():
# Take input Here and Call solution function
for _ in range(get_int()):
n = get_int()
arr = get_ints_in_list()
Solution(arr, n)
# calling main Function
if __name__ == '__main__':
main()
``` | output | 1 | 54,204 | 1 | 108,409 |
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