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）

  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\n", ans);
102     }
103     return 0;
104 }