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）
1 #include <cstdio> 2 #include <algorithm> 3 #include <iostream> 4 #include <cstring> 5 #include <string> 6 #include <cmath> 7 #include <queue> 8 #include <vector> 9 #include <stack> 10 using namespace std; 11 #define N 20010 12 #define M 50010 13 struct node 14 { 15 int v, next, u; 16 }edge[M]; 17 int n, tot, cnt, num, head[N], dfn[N], low[N], belong[N], in[N], out[N]; 18 bool vis[N]; 19 stack<int> sta; 20 21 void init() 22 { 23 tot = 0; 24 num = 0; 25 cnt = 0; 26 while(!sta.empty()) sta.pop(); 27 memset(head, -1, sizeof(head)); 28 memset(vis, false, sizeof(vis)); 29 memset(low, 0, sizeof(low)); 30 memset(dfn, 0, sizeof(dfn)); 31 memset(belong, 0, sizeof(belong)); 32 memset(in, 0, sizeof(in)); 33 memset(out, 0, sizeof(out)); 34 } 35 36 void add(int u, int v) 37 { 38 edge[tot].next = head[u]; edge[tot].v = v; edge[tot].u = u; head[u] = tot++; 39 } 40 41 void tarjan(int u) 42 { 43 dfn[u] = low[u] = ++cnt; 44 sta.push(u); 45 vis[u] = true; 46 for(int i = head[u]; ~i; i = edge[i].next) { 47 int v = edge[i].v; 48 if(!dfn[v]) { 49 tarjan(v); 50 if(low[v] < low[u]) low[u] = low[v]; 51 } else { 52 if(vis[v] && low[u] > dfn[v]) low[u] = dfn[v]; 53 } 54 } 55 if(dfn[u] == low[u]) { 56 num++; 57 int top = 0; 58 while(top != u) { 59 top = sta.top(); 60 belong[top] = num; 61 vis[top] = 0; 62 sta.pop(); 63 } 64 } 65 } 66 67 int main() 68 { 69 int t; 70 scanf("%d", &t); 71 while(t--) { 72 init(); 73 int m; 74 scanf("%d%d", &n, &m); 75 for(int i = 0; i < m; i++) { 76 int u, v; 77 scanf("%d%d", &u, &v); 78 add(u, v); 79 } 80 for(int i = 1; i <= n; i++) { 81 if(!dfn[i]) { 82 tarjan(i); 83 } 84 } 85 for(int u = 1; u <= n; u++) { 86 for(int i = head[u]; ~i; i = edge[i].next) { 87 int v = edge[i].v; 88 if(belong[u] != belong[v]) { 89 in[belong[u]]++; 90 out[belong[v]]++; 91 } 92 } 93 } 94 int inn = 0, outt = 0; 95 for(int i = 1; i <= num; i++) { 96 if(in[i] == 0) inn++; 97 if(out[i] == 0) outt++; 98 } 99 int ans = max(inn, outt); 100 if(num == 1) ans = 0; 101 printf("%d ", ans); 102 } 103 return 0; 104 }