loj 1210 (求最少的加边数使得图变成强连通)

题目链接：http://lightoj.com/volume_showproblem.php?problem=1210

思路：首先是缩点染色，然后重建并且统计新图中的每个点的入度和出度，于是答案就是max(入度为0的点的个数， 出度为0的点的个数，这里有一个trick就是如果scc_count == 1，那么应该输出0.

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <stack>
 using namespace std;

 const int MAXN = ( + );
 int n, m, NE;
 struct Edge {
     int v, next;
 } edge[MAXN << ];

 int head[MAXN];
 void Insert(int u, int v)
 {
     edge[NE].v = v;
     edge[NE].next = head[u];
     head[u] = NE++;
 }

 int cnt, scc_count;
 int low[MAXN], dfn[MAXN], color[MAXN];
 bool mark[MAXN];
 stack<int >S;

 void Tarjan(int u)
 {
     low[u] = dfn[u] = ++cnt;
     mark[u] = true;
     S.push(u);
     for (int i = head[u]; i != -; i = edge[i].next) {
         int v = edge[i].v;
         if (dfn[v] == ) {
             Tarjan(v);
             low[u] = min(low[u], low[v]);
         } else if (mark[v]) {
             low[u] = min(low[u], dfn[v]);
         }
     }
     if (low[u] == dfn[u]) {
         scc_count++;
         int v;
         do {
             v = S.top();
             S.pop();
             mark[v] = false;
             color[v] = scc_count;
         } while (u != v);
     }
 }

 int Indegree[MAXN], Outdegree[MAXN];

 int main()
 {
     int _case, t = ;
     scanf("%d", &_case);
     while (_case--) {
         scanf("%d %d", &n, &m);
         NE = ;
         memset(head, -, sizeof(head));
         while (m--) {
             int u, v;
             scanf("%d %d", &u, &v);
             Insert(u, v);
         }
         cnt = scc_count = ;
         memset(dfn, , sizeof(dfn));
         memset(mark, false, sizeof(mark));
         for (int i = ; i <= n; i++) {
             if (dfn[i] == ) Tarjan(i);
         }
         memset(Indegree, , sizeof(Indegree));
         memset(Outdegree, , sizeof(Outdegree));
         for (int u = ; u <= n; u++) {
             for (int i = head[u]; i != -; i = edge[i].next) {
                 int v = edge[i].v;
                 if (color[u] != color[v]) {
                     Indegree[color[v]]++;
                     Outdegree[color[u]]++;
                 }
             }
         }
         int ans1 = , ans2 = ;
         for (int i = ; i <= scc_count; i++) {
             if (Indegree[i] == ) ans1++;
             if (Outdegree[i] == ) ans2++;
         }
         printf("Case %d: ", t++);
         if (scc_count == ) {
             puts("");
         } else
             printf("%d\n", max(ans1, ans2));
     }
     return ;
 }