hdu3861The King’s Problem

题目意思就是一个国家有n个城市，里面的道路都是单向的，现在国王要把国家分割成区域，要求的就是，每个区域里面的点u、v，u到v可达或者是v到u可达。求最少要分隔成几个区域。

显然就是最小路径覆盖问题，只是要缩点。

/*****************************************Author      :Crazy_AC(JamesQi)Time        :2015File Name   :*****************************************/// #pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <algorithm>#include <iomanip>#include <sstream>#include <string>#include <stack>#include <queue>#include <deque>#include <vector>#include <map>#include <set>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <climits>using namespace std;#define MEM(x,y) memset(x, y,sizeof x)#define pk push_backtypedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> ii;typedef pair<ii,int> iii;const double eps = 1e-10;const int inf = 1 << 30;const int INF = 0x3f3f3f3f;const int N = 5000 + 10;const int M = 100000 + 10;int n, m;int head[N], pnt[M], nxt[M], cnt;void init() {memset(head, -1,sizeof head);cnt = 0;}void addedge(int u,int v) {pnt[cnt] = v, nxt[cnt] = head[u],head[u] = cnt++;}int dfn[N], low[N], belong[N];int Times;stack<int> st;int scc;void Tarjan(int u) {dfn[u] = low[u] = ++Times;st.push(u);for (int i = head[u];~i;i = nxt[i]) {int v = pnt[i];if (!dfn[v]) {Tarjan(v);low[u] = min(low[u], low[v]);}else if (!belong[v]) low[u] = min(low[u], dfn[v]);}if (dfn[u] == low[u]) {++scc;while(true) {int x = st.top();st.pop();belong[x] = scc;if (x == u) break;}}}vector<int> G[N];int linker[N];bool vis[N];bool dfs(int u) {for (int i = 0;i < G[u].size();++i) {int v = G[u][i];if (vis[v]) continue;vis[v] = true;if (linker[v] == -1 || dfs(linker[v])) {linker[v] = u;return true;}}return false;}int Hungary(){int ans = 0;memset(linker, -1,sizeof linker);for (int i = 1;i <= scc;++i) {memset(vis, false,sizeof vis);if (dfs(i)) ans++;}return ans;}int main(){// freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);int t;scanf("%d",&t);while(t--) {scanf("%d%d",&n,&m);init();int u, v;for (int i = 1;i <= m;++i) {scanf("%d%d",&u,&v);addedge(u, v);}memset(dfn, 0,sizeof dfn);memset(belong, 0,sizeof belong);scc = Times = 0;for (int i = 1;i <= n;++i) G[i].clear();for (int i = 1;i <= n;++i) if (!dfn[i]) Tarjan(i);for (int i = 1;i <= n;++i) {for (int j = head[i];~j;j = nxt[j]) {int v = pnt[j];if (belong[i] != belong[v]) {G[belong[i]].push_back(belong[v]);}}}printf("%d\n", scc - Hungary());}return 0;}