HDU 4635：Strongly connected（强连通）

http://acm.hdu.edu.cn/showproblem.php?pid=4635

题意：给出n个点和m条边，问最多能添加几条边使得图不是一个强连通图。如果一开始强连通就-1.思路：把图分成x个强连通分量之后，每个强连通分量最大的边数是n*（n-1），然后考虑和其他强连通分量相连的情况：即把分量a的所有点都连向分量b的所有点，而b不连a，这样就可以让图不是强连通的。可以把整个图分成两个强连通分量a和b分别有i和j个点，其中i+j=n，那么答案就是n*（n-1）-m-i*j。所以求出最小的i*j就可以找到答案了。

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <cstring>
 #include <string>
 #include <cmath>
 #include <queue>
 #include <vector>
 #include <stack>
 using namespace std;
 #define N 100010
 struct node
 {
     int v, next, u;
 }edge[N];

 int n, tot, cnt, num, head[N], dfn[N], low[N], belong[N], in[N], out[N], deg[N], tol[N], e[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));
     memset(deg, , sizeof(deg));
     memset(tol, , sizeof(tol));
     memset(e, , sizeof(e));
 }

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

 void tarjan(int u)
 {
     vis[u] = ; sta.push(u);
     dfn[u] = low[u] = ++cnt;
     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]) {
             if(dfn[v] < low[u]) low[u] = dfn[v];
         }
     }
     if(low[u] == dfn[u]) {
         ++num;
         int top = -;
         while(top != u) {
             top = sta.top(); sta.pop();
             vis[top] = ;
             belong[top] = num;
         }
     }
 }

 bool cmp(const int &a, const int &b)
 {
     if(out[a] != ) return false;
     if(out[b] != ) return true;
     return tol[a] < tol[b];
 }

 int main()
 {
     int t, cas = ;
     scanf("%d", &t);
     while(t--) {
         int n, m;
         scanf("%d%d", &n, &m);
         init();
         for(int i = ; i < m; i++) {
             int u, v;
             scanf("%d%d", &u, &v);
             add(u, v);
             deg[u]++; deg[v]++;
         }
         for(int i = ; i <= n; i++) {
             if(!dfn[i]) tarjan(i);
         }
         printf("Case %d: ", cas++);
         if(num == ) {
             puts("-1"); continue;
         }
 //        printf("~~~\n");
         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[v]]++;
                     out[belong[u]]++;
                 }
             }
         }
         long long sum = (long long)n * (n - ) - m;
         for(int i = ; i <= n; i++) {
             int tmp = belong[i];
             tol[tmp]++;
         }
         long long ans = ;
         for(int i = ; i <= num; i++) {
             if(!in[i] || !out[i]) {
                 ans = max(ans, sum - (long long)(n - tol[i]) * tol[i]);
             }
         }
         printf("%I64d\n", ans);
     }
     return ;
 }

 /*
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 */