hdu4635Strongly connected

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

tarjan缩点 统计缩点后每个结点的出度入度 将那个包含原来点数最少的 且出度或者入度为0的大节点看作一个整体内部连边n*（n-1)个 连全部的； 其它的点为一整体连全部的 再两者连一同向的边 保证它的出度或者入度依旧为0的情况下任意连 最后减去原来的边M

 #include <iostream>
 #include<cstring>
 #include<cstdio>
 #include<stdlib.h>
 #include<stack>
 using namespace std;
 #define N 100010
 #define M 100010
 #define INF 0xfffffff
 int a[][];
 stack<int>s;
 struct node
 {
     int u,v,next;
 }ed[M*];
 int scc,sccno[N],head[N],lowlink[N],pre[N],dep,num[N],x[M],y[M],din[N],dout[N],o;
 void init()
 {
     o = ;
     memset(head,-,sizeof(head));
 }
 void add(int u,int v)
 {
     ed[o].u = u;
     ed[o].v = v;
     ed[o].next = head[u];
     head[u] = o++;
 }
 void dfs(int u)
 {
     int i;
     lowlink[u] = pre[u] = ++dep;
     s.push(u);
     for( i = head[u] ; i != - ; i = ed[i].next)
     {
         int v = ed[i].v;
         if(!pre[v])
         {
             dfs(v);
             lowlink[u] = min(lowlink[u],lowlink[v]);
         }
         else if(!sccno[v])
         lowlink[u] = min(lowlink[u],pre[v]);
     }
     if(lowlink[u]==pre[u])
     {
         scc++;
         for(;;)
         {
             int x = s.top();
             s.pop();
             sccno[x] = scc;
             if(x==u) break;
         }
     }
 }
 void find_scc(int n)
 {
     for(int i = ; i <= n ;i++)
     if(!pre[i])
     dfs(i);
 }
 int main()
 {
     int i,t,n,m,a,b,kk=;
     cin>>t;
     while(t--)
     {
         init();kk++;
         memset(num,,sizeof(num));
         memset(pre,,sizeof(pre));
         memset(din,,sizeof(din));
         memset(dout,,sizeof(dout));
         memset(sccno,,sizeof(sccno));
         scanf("%d%d",&n,&m);
         for(i = ; i <= m ; i++)
         {
             scanf("%d%d",&a,&b);
             x[i] = a;y[i] = b;
             add(a,b);
         }
         scc=;dep=;
         find_scc(n);
         printf("Case %d: ",kk);
         if(scc==)
         {
             cout<<"-1\n";
             continue;
         }
         for(i = ; i <= n ; i++)
         num[sccno[i]]++;
         for(i = ; i <= m ;i++)
         {
             if(sccno[x[i]]!=sccno[y[i]])
             {
                 dout[sccno[x[i]]]++;
                 din[sccno[y[i]]]++;
             }
         }
         long long minz = INF;
         for(i =  ;i <= scc ; i++)
         if(minz>num[i]&&(din[i]==||dout[i]==))
         minz = num[i];
         long long ss=;
         ss = minz*(minz-)+(n-minz)*(n-minz-)+minz*(n-minz);
         cout<<ss-m<<endl;
     }
     return ;
 }