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
1 #include <iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<stdlib.h> 5 #include<stack> 6 using namespace std; 7 #define N 100010 8 #define M 100010 9 #define INF 0xfffffff 10 int a[110][110]; 11 stack<int>s; 12 struct node 13 { 14 int u,v,next; 15 }ed[M*2]; 16 int scc,sccno[N],head[N],lowlink[N],pre[N],dep,num[N],x[M],y[M],din[N],dout[N],o; 17 void init() 18 { 19 o = 0; 20 memset(head,-1,sizeof(head)); 21 } 22 void add(int u,int v) 23 { 24 ed[o].u = u; 25 ed[o].v = v; 26 ed[o].next = head[u]; 27 head[u] = o++; 28 } 29 void dfs(int u) 30 { 31 int i; 32 lowlink[u] = pre[u] = ++dep; 33 s.push(u); 34 for( i = head[u] ; i != -1 ; i = ed[i].next) 35 { 36 int v = ed[i].v; 37 if(!pre[v]) 38 { 39 dfs(v); 40 lowlink[u] = min(lowlink[u],lowlink[v]); 41 } 42 else if(!sccno[v]) 43 lowlink[u] = min(lowlink[u],pre[v]); 44 } 45 if(lowlink[u]==pre[u]) 46 { 47 scc++; 48 for(;;) 49 { 50 int x = s.top(); 51 s.pop(); 52 sccno[x] = scc; 53 if(x==u) break; 54 } 55 } 56 } 57 void find_scc(int n) 58 { 59 for(int i = 1; i <= n ;i++) 60 if(!pre[i]) 61 dfs(i); 62 } 63 int main() 64 { 65 int i,t,n,m,a,b,kk=0; 66 cin>>t; 67 while(t--) 68 { 69 init();kk++; 70 memset(num,0,sizeof(num)); 71 memset(pre,0,sizeof(pre)); 72 memset(din,0,sizeof(din)); 73 memset(dout,0,sizeof(dout)); 74 memset(sccno,0,sizeof(sccno)); 75 scanf("%d%d",&n,&m); 76 for(i = 1; i <= m ; i++) 77 { 78 scanf("%d%d",&a,&b); 79 x[i] = a;y[i] = b; 80 add(a,b); 81 } 82 scc=0;dep=0; 83 find_scc(n); 84 printf("Case %d: ",kk); 85 if(scc==1) 86 { 87 cout<<"-1 "; 88 continue; 89 } 90 for(i = 1; i <= n ; i++) 91 num[sccno[i]]++; 92 for(i = 1; i <= m ;i++) 93 { 94 if(sccno[x[i]]!=sccno[y[i]]) 95 { 96 dout[sccno[x[i]]]++; 97 din[sccno[y[i]]]++; 98 } 99 } 100 long long minz = INF; 101 for(i = 1 ;i <= scc ; i++) 102 if(minz>num[i]&&(din[i]==0||dout[i]==0)) 103 minz = num[i]; 104 long long ss=0; 105 ss = minz*(minz-1)+(n-minz)*(n-minz-1)+minz*(n-minz); 106 cout<<ss-m<<endl; 107 } 108 return 0; 109 }