思路:
题目标签是记忆化搜索,但是这种题我用的拓扑排序
首先建好图,将多余的边删掉,再从A点开始拓扑到B点,如果可以推到其他无出边的点,则不是逻辑自洽。路上记录边数即可求出A到B的条数。
Tip:
拓扑排序注意拓扑到B点就立刻停止,如果WA了测试点4极有可能是这种情况。
#include <bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 1000 + 5; const int INF = 0x7FFFFFFF; vector<int> chu[maxn]; int ru[maxn]; int cnt[maxn]; int main() { int n, m; cin >> n >> m; for (int i = 1; i <= m; i++) { int x, y; cin >> x >> y; chu[x].push_back(y); ru[y]++; } int A, B; cin >> A >> B; queue<int> que2; for (int i = 0; i <= maxn; i++) { if (ru[i] == 0 && i != A) que2.push(i); } while (!que2.empty()) { int now = que2.front(); que2.pop(); for (auto i:chu[now]) { ru[i]--; if (ru[i] == 0 && i != A) que2.push(i); } } queue<int> que; que.push(A); cnt[A] = 1; bool flag1 = true; while (!que.empty()) { int now = que.front(); que.pop(); if (now == B) continue; if (chu[now].empty() && now != B) flag1 = false; for (auto i:chu[now]) { cnt[i] += cnt[now]; ru[i]--; if (ru[i] == 0) que.push(i); } } cout << cnt[B] << " "; if (flag1) cout << "Yes" << endl; else cout << "No" << endl; return 0; }