HDU 2767：Proving Equivalences（强连通）

http://acm.hdu.edu.cn/showproblem.php?pid=2767

题意：给出n个点m条边，问在m条边的基础上，最小再添加多少条边可以让图变成强连通。思路：强连通分量缩点后找入度为0和出度为0的点，因为在强连通图里面没有一个点的入度和出度都为0，所以取出度为0的点和入度为0的点中的最大值就是答案。（要特判强连通分量数为1的情况）

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <cstring>
 #include <string>
 #include <cmath>
 #include <queue>
 #include <vector>
 #include <stack>
 using namespace std;
 #define N 20010
 #define M 50010
 struct node
 {
     int v, next, u;
 }edge[M];
 int n, tot, cnt, num, head[N], dfn[N], low[N], belong[N], in[N], out[N];
 bool vis[N];
 stack<int> sta;

 void init()
 {
     tot = ;
     num = ;
     cnt = ;
     while(!sta.empty()) sta.pop();
     memset(head, -, sizeof(head));
     memset(vis, false, sizeof(vis));
     memset(low, , sizeof(low));
     memset(dfn, , sizeof(dfn));
     memset(belong, , sizeof(belong));
     memset(in, , sizeof(in));
     memset(out, , sizeof(out));
 }

 void add(int u, int v)
 {
     edge[tot].next = head[u]; edge[tot].v = v; edge[tot].u = u; head[u] = tot++;
 }

 void tarjan(int u)
 {
     dfn[u] = low[u] = ++cnt;
     sta.push(u);
     vis[u] = true;
     for(int i = head[u]; ~i; i = edge[i].next) {
         int v = edge[i].v;
         if(!dfn[v]) {
             tarjan(v);
             if(low[v] < low[u]) low[u] = low[v];
         } else {
             if(vis[v] && low[u] > dfn[v]) low[u] = dfn[v];
         }
     }
     if(dfn[u] == low[u]) {
         num++;
         int top = ;
         while(top != u) {
             top = sta.top();
             belong[top] = num;
             vis[top] = ;
             sta.pop();
         }
     }
 }

 int main()
 {
     int t;
     scanf("%d", &t);
     while(t--) {
         init();
         int m;
         scanf("%d%d", &n, &m);
         for(int i = ; i < m; i++) {
             int u, v;
             scanf("%d%d", &u, &v);
             add(u, v);
         }
         for(int i = ; i <= n; i++) {
             if(!dfn[i]) {
                 tarjan(i);
             }
         }
         for(int u = ; u <= n; u++) {
             for(int i = head[u]; ~i; i = edge[i].next) {
                 int v = edge[i].v;
                 if(belong[u] != belong[v]) {
                     in[belong[u]]++;
                     out[belong[v]]++;
                 }
             }
         }
         int inn = , outt = ;
         for(int i = ; i <= num; i++) {
             if(in[i] == ) inn++;
             if(out[i] == ) outt++;
         }
         int ans = max(inn, outt);
         if(num == ) ans = ;
         printf("%d\n", ans);
     }
     return ;
 }