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题意:有向无环图,求最少多少条路径能够覆盖整个图,点能够反复走
思路:和普通的最小路径覆盖不同的是,点能够反复走,那么事实上仅仅要在多一步。利用floyd求出传递闭包。然后依据这个新的图去做最小路径覆盖就可以
代码:
#include <cstdio> #include <cstring> #include <vector> #include <algorithm> using namespace std; const int N = 505; int n, m, g[N][N]; int left[N], vis[N]; bool dfs(int u) { for (int i = 1; i <= n; i++) { if (g[u][i] && !vis[i]) { vis[i] = 1; if (!left[i] || dfs(left[i])) { left[i] = u; return true; } } } return false; } int hungary() { int ans = 0; memset(left, 0, sizeof(left)); for (int i = 1; i <= n; i++) { memset(vis, 0, sizeof(vis)); if (dfs(i)) ans++; } return ans; } int main() { while (~scanf("%d%d", &n, &m) && n) { int u, v; memset(g, 0, sizeof(g)); while (m--) { scanf("%d%d", &u, &v); g[u][v] = 1; } for (int k = 1; k <= n; k++) { for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { g[i][j] |= (g[i][k]&g[k][j]); } } } printf("%d\n", n - hungary()); } return 0; }
POJ 2594 Treasure Exploration(最小路径覆盖变形)
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原文地址:http://www.cnblogs.com/mengfanrong/p/5184910.html