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problems/PA-01/Atcoder/abc177_f/editorial.md

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+# F - I hate Shortest Path Problem (editorial)
+
+Source: https://atcoder.jp/contests/abc177/editorial/94 by evima
+
+When \(k\) is fixed, every path has the same number \(k\) of moving downwards, so we want to find the number of minimum times to move to the right.
+
+The optimal strategy of moving is to move downwards as much as possible, and if not possible, move to the right. This is due to that fact that, since we want to minimize the number of movement towards right, we can always move to the right in the same way in every row, thus we do not have to move to the right until needed. Now, the path is uniquely determined when the starting square is fixed.
+
+Calculate in the increasing order of \(k\) while managing the possible \(W\) pairs of positions of stariting end ending squares counted from the leftmost one. In the \(k\)-th process, for each pair such that its ending square is between \(A_k\) and \(B_k\), inclusive, upadte its ending square to \(B_k+1\), then answer the minimum difference of starting and ending squares. Note that however, when \(B_k = W\), the endpoint should be updated to \(\infty\) instead of \(W+1\).
+
+This operations can be processed with semgnet trees, but this can also be processed with ordered sets.
+
+Once the ending square of two pairs becomes the same, they will have the common endpoint from then on, so all of them except for the one having the maximum starting point can be ignored. Prepare a set of (ending point, staring point) and a multiset of endting points subtracted by starting points, and update them while removing the elements that are no longer needed. The total time complexity will be \(O((H+W)log(H+W))\).

problems/PA-01/Atcoder/abc177_f/judge.json

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+{
+  "time_limit_s": 2.0,
+  "time_limit_source": "task page",
+  "memory_mb": 1024,
+  "memory_source": "task page",
+  "mode": "exact",
+  "validator_flags": "case_sensitive",
+  "comparison": "token-wise, whitespace-insensitive, case-sensitive (AtCoder's checker)",
+  "validator": null,
+  "tests_dir": "data",
+  "answer_suffix": ".out",
+  "n_sample": null,
+  "n_secret": 11,
+  "note": "no sample/secret split: the statement's samples are among the numbered tests"
+}

problems/PA-01/Atcoder/abc177_f/meta.json

@@ -0,0 +1,16 @@
+{
+  "task_id": "abc177_f",
+  "contest": "abc177",
+  "type": "ABC",
+  "letter": "F",
+  "title": "I hate Shortest Path Problem",
+  "url": "https://atcoder.jp/contests/abc177/tasks/abc177_f",
+  "difficulty": 2291,
+  "score": 600,
+  "time_limit_sec": 2.0,
+  "memory_limit_mb": 1024,
+  "n_tests": 11,
+  "editorial_url": "https://atcoder.jp/contests/abc177/editorial/94",
+  "code_source": "model",
+  "pid": "A1440"
+}

problems/PA-01/Atcoder/abc177_f/solution.cpp

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+// abc177_f: F - I hate Shortest Path Problem
+// Written by a model (gpt-6-astra) from the official editorial, https://atcoder.jp/contests/abc177/editorial/94
+// Judged AC on all 11 tests, max 0.16 s of the 2 s limit (submission 1).
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int H, W;
+    cin >> H >> W;
+
+    map<int, int> states; // endpoint -> largest starting column
+    multiset<int> costs;  // endpoint - starting column
+    for (int j = 1; j <= W; ++j) {
+        states.emplace_hint(states.end(), j, j);
+        costs.insert(costs.end(), 0);
+    }
+
+    for (int row = 1; row <= H; ++row) {
+        int a, b;
+        cin >> a >> b;
+
+        auto it = states.lower_bound(a);
+        int best_start = -1;
+        while (it != states.end() && it->first <= b) {
+            best_start = max(best_start, it->second);
+            costs.erase(costs.find(it->first - it->second));
+            it = states.erase(it);
+        }
+
+        if (best_start != -1 && b < W) {
+            int endpoint = b + 1;
+            if (it != states.end() && it->first == endpoint) {
+                if (best_start > it->second) {
+                    costs.erase(costs.find(endpoint - it->second));
+                    it->second = best_start;
+                    costs.insert(endpoint - best_start);
+                }
+            } else {
+                states.emplace_hint(it, endpoint, best_start);
+                costs.insert(endpoint - best_start);
+            }
+        }
+
+        if (costs.empty()) {
+            cout << -1 << '\n';
+        } else {
+            cout << row + *costs.begin() << '\n';
+        }
+    }
+    return 0;
+}

problems/PA-01/Atcoder/abc177_f/statement.md

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+# F - I hate Shortest Path Problem
+
+**Score:** 600 points
+
+## Problem Statement
+There is a grid of squares with \(H+1\) horizontal rows and \(W\) vertical columns.  
+You will start at one of the squares in the top row and repeat moving one square right or down. However, for each integer \(i\) from \(1\) through \(H\), you cannot move down from the \(A_i\)-th, \((A_i + 1)\)-th, \(\ldots\), \(B_i\)-th squares from the left in the \(i\)-th row from the top.  
+For each integer \(k\) from \(1\) through \(H\), find the minimum number of moves needed to reach one of the squares in the \((k+1)\)-th row from the top. (The starting square can be chosen individually for each case.) If, starting from any square in the top row, none of the squares in the \((k+1)\)-th row can be reached, print `-1` instead.
+
+### Constraints
+- \(1 \leq H,W \leq 2\times 10^5\)
+- \(1 \leq A_i \leq B_i \leq W\)
+
+---
+
+### Input
+Input is given from Standard Input in the following format:
+```
+H W
+A_1 B_1
+A_2 B_2
+:
+A_H B_H
+```
+
+### Output
+Print \(H\) lines. The \(i\)-th line should contain the answer for the case \(k=i\).
+
+---
+
+### Sample Input 1
+```
+4 4
+2 4
+1 1
+2 3
+2 4
+```
+
+### Sample Output 1
+```
+1
+3
+6
+-1
+```
+
+Let \((i,j)\) denote the square at the \(i\)-th row from the top and \(j\)-th column from the left.  
+For \(k=1\), we need one move such as \((1,1) \to (2,1)\).  
+For \(k=2\), we need three moves such as \((1,1) \to (2,1) \to (2,2) \to (3,2)\).  
+For \(k=3\), we need six moves such as \((1,1) \to (2,1) \to (2,2) \to (3,2) \to (3,3) \to (3,4) \to (4,4)\).  
+For \(k=4\), it is impossible to reach any square in the fifth row from the top.

problems/PA-01/Atcoder/abc225_e/editorial.md

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+# E - 7 (editorial)
+
+Source: https://atcoder.jp/contests/abc225/editorial/2881 by en_translator
+
+First of all, after removing some 7’s, if there still remain 7’s that are not wholly visible from the origin, then we can remove such 7’s as well without changing the number of 7’s that are wholly visible from the origin.
+
+Therefore, the problem can be rephrased to as follows:
+
+Given \(N\) number of 7’s, you may remove some of them so that all remaining 7’s are visible from the origin. Find the maximum number of remaining 7’s.
+
+Moreover, one can regard each 7 as a open interval corresponding to the polar angles of its both endpoints, so that the problem is reworded as follows:
+
+Given are \(N\) open intervals. The left endpoint of the \(i\)-th interval is \(f(x_i,y_i-1)\) and that of the right endpoint is \(f(x_i-1,y_i)\), where \(f(i,j)\) is a function of two real numbers \((i,j)\) that represents the polar angle of coordinate \((i,j)\).
+
+At most how many intervals can be chosen from them, so that no two different intervals do not overlap?
+
+This is equivalent to the famous problem called interval scheduling problem. It is very famous that the interval scheduling problem can be solved by the following greedy algorithm, which can be applied for this problem as well.
+
+Repeat the following operations infinitely.
+
+1. Let \(R_{\max}\) be the maximum coordinate of (the right endpoint of) already chosen intervals. If there exist not yet chosen intervals with the leftmost coordinates \(R_{\max}\) or greater, then choose one with the smallest rightmost coordinate.
+2. Otherwise, if there are no such interval with the leftmost coordinate greater than or equal to \(R_{\max}\), then terminate the procedure.
+
+When implementing, it is a good idea to sort \(N\) intervals in the increasing order of their rightmost coordinates. The complexity is \(O(N \log N)\).

problems/PA-01/Atcoder/abc225_e/judge.json

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+{
+  "time_limit_s": 2.0,
+  "time_limit_source": "task page",
+  "memory_mb": 1024,
+  "memory_source": "task page",
+  "mode": "exact",
+  "validator_flags": "case_sensitive",
+  "comparison": "token-wise, whitespace-insensitive, case-sensitive (AtCoder's checker)",
+  "validator": null,
+  "tests_dir": "data",
+  "answer_suffix": ".out",
+  "n_sample": null,
+  "n_secret": 39,
+  "note": "no sample/secret split: the statement's samples are among the numbered tests"
+}

problems/PA-01/Atcoder/abc225_e/meta.json

@@ -0,0 +1,16 @@
+{
+  "task_id": "abc225_e",
+  "contest": "abc225",
+  "type": "ABC",
+  "letter": "E",
+  "title": "7",
+  "url": "https://atcoder.jp/contests/abc225/tasks/abc225_e",
+  "difficulty": 1678,
+  "score": 500,
+  "time_limit_sec": 2.0,
+  "memory_limit_mb": 1024,
+  "n_tests": 39,
+  "editorial_url": "https://atcoder.jp/contests/abc225/editorial/2881",
+  "code_source": "editorial",
+  "pid": "A1753"
+}

problems/PA-01/Atcoder/abc225_e/solution.cpp

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+// abc225_e: E - 7
+// Code from the official editorial, https://atcoder.jp/contests/abc225/editorial/2881 (by en_translator)
+// Judged AC on all 39 tests, max 0.10 s of the 2 s limit.
+
+#include<bits/stdc++.h>
+using namespace std;
+
+using ll = long long;
+
+// p/q
+struct fraction{
+    ll p,q;
+    fraction(ll P = 0, ll Q = 1): p(P), q(Q){}
+    bool operator<(const fraction &other)const{
+        return p*other.q < other.p*q;
+    }
+    bool operator<=(const fraction &other)const{
+        return p*other.q <= other.p*q;
+    }
+};
+
+int main(){
+    int N; cin >> N;
+    vector<ll> x(N),y(N);
+    for(int i=0; i<N; i++) cin >> x[i] >> y[i];
+    vector<pair<fraction,fraction>> data(N);
+    for(int i=0; i<N; i++) data[i] = make_pair(fraction(y[i],x[i]-1),fraction(y[i]-1,x[i]));
+    sort(data.begin(),data.end());
+    int ans = 0;
+    fraction now;
+    for(auto p: data){
+        if(now <= p.second){
+            ans++;
+            now = p.first;
+        }
+    }
+    cout << ans << endl;
+}

problems/PA-01/Atcoder/abc225_e/statement.md

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+# E - 7
+
+**Score:** 500 points
+
+## Problem Statement
+There are \(N\) 7's drawn in the first quadrant of a plane.  
+The \(i\)-th 7 is a figure composed of a segment connecting \((x_i-1,y_i)\) and \((x_i,y_i)\), and a segment connecting \((x_i,y_i-1)\) and \((x_i,y_i)\).  
+You can choose zero or more from the \(N\) 7's and delete them.  
+What is the maximum possible number of 7's that are wholly visible from the origin after the optimal deletion?  
+Here, the \(i\)-th 7 is wholly visible from the origin if and only if:
+- the interior (excluding borders) of the quadrilateral whose vertices are the origin, \((x_i-1,y_i)\), \((x_i,y_i)\), \((x_i,y_i-1)\) does not intersect with the other 7's.
+
+### Constraints
+- \(1 \leq N \leq 2 \times 10^5\)
+- \(1 \leq x_i,y_i \leq 10^9\)
+- \((x_i,y_i) \neq (x_j,y_j)\ (i \neq j)\)
+- All values in input are integers.
+
+---
+
+## Input
+Input is given from Standard Input in the following format:
+```
+N
+x_1 y_1
+x_2 y_2
+...
+x_N y_N
+```
+
+## Output
+Print the maximum possible number of 7's that are wholly visible from the origin.
+
+---
+
+## Sample Input 1
+```
+3
+1 1
+2 1
+1 2
+```
+
+## Sample Output 1
+```
+2
+```
+If the first 7 is deleted, the other two 7's ― the second and third ones ― will be wholly visible from the origin, which is optimal.  
+If no 7's are deleted, only the first 7 is wholly visible from the origin.
+
+---
+
+## Sample Input 2
+```
+10
+414598724 87552841
+252911401 309688555
+623249116 421714323
+605059493 227199170
+410455266 373748111
+861647548 916369023
+527772558 682124751
+356101507 249887028
+292258775 110762985
+850583108 796044319
+```
+
+## Sample Output 2
+```
+10
+```
+It is best to keep all 7's.

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