题目链接：

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

Strongly connected

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Give a simple directed graph with N nodes and M edges. Please tell me the maximum number of the edges you can add that the graph is still a simple directed graph. Also, after you add these edges, this graph must NOT be strongly connected.

A simple directed graph is a directed graph having no multiple edges or graph loops.

Input

The first line of date is an integer T, which is the number of the text cases.

Then T cases follow, each case starts of two numbers N and M, 1<=N<=100000, 1<=M<=100000, representing the number of nodes and the number of edges, then M lines follow. Each line contains two integers x and y, means that there is a edge from x to y.

Output

For each case, you should output the maximum number of the edges you can add.

If the original graph is strongly connected, just output -1.

Sample Input

3
3 3
1 2
2 3
3 1
3 3
1 2
2 3
1 3
6 6
1 2
2 3
3 1
4 5
5 6
6 4

Sample Output

Case 1: -1
Case 2: 1
Case 3: 15

Source

2013 Multi-University Training Contest 4

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题目意思：

给一幅图，求最多增加多少边，可以使该图能够不满足强连通性。

解题思路：

分析知，最后肯定是两个完全图a,b，然后a中每个节点到b中每个节点都有一条有向边。整体考虑总的边数为a*(a-1)+b*(b-1)+a*b=n*n-n-a*b.

ans=n*n-n-a*b-m  n/m已知，所以让ans最大，就需让a*b最小，使得a,b差距最大。所以找最小的a或b.

tarjan求强连通，然后求出入度或出度为0的强连通分量，统计这些联通分量中节点数最小的a.

代码：

//#include<CSpreadSheet.h>

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

#define Maxn 110000

int low[Maxn],dfn[Maxn],sta[Maxn],bc,sc,dep;
ll n,m;
int in[Maxn],dei[Maxn],deo[Maxn],nu[Maxn];
vector<vector<int> >myv;
bool iss[Maxn];

void tarjan(int cur)
{
    int ne;

    low[cur]=dfn[cur]=++dep;
    sta[++sc]=cur;
    iss[cur]=true;
    for(int i=0;i<myv[cur].size();i++)
    {
        ne=myv[cur][i];
        if(!dfn[ne])
        {
            tarjan(ne);
            if(low[ne]<low[cur])
                low[cur]=low[ne];
        }
        else if(iss[ne]&&dfn[ne]<low[cur])
            low[cur]=dfn[ne];
    }
    if(low[cur]==dfn[cur])
    {
        bc++;
        do
        {
            ne=sta[sc--];
            iss[ne]=false;
            in[ne]=bc;
            nu[bc]++;
        }while(ne!=cur);
    }
}
void solve()
{
    dep=sc=bc=0;
    memset(nu,0,sizeof(nu));
    memset(iss,false,sizeof(iss));
    memset(dfn,0,sizeof(dfn));
    for(int i=1;i<=n;i++)
        if(!dfn[i])
            tarjan(i);
}

int main()
{
    //freopen("in.txt","r",stdin);
   //freopen("out.txt","w",stdout);
   int t,kcas=0;

   scanf("%d",&t);
   while(t--)
   {
       scanf("%I64d%I64d",&n,&m);
       myv.clear();
       myv.resize(n+1);

       for(int i=1;i<=m;i++)
       {
           int a,b;
           scanf("%d%d",&a,&b);
           myv[a].push_back(b);
       }
       solve();
       if(bc==1)
       {
           printf("Case %d: -1\n",++kcas);
           continue;
       }
       memset(dei,0,sizeof(dei));
       memset(deo,0,sizeof(deo));

       for(int i=1;i<=n;i++)
            for(int j=0;j<myv[i].size();j++)
            {
               int ne=myv[i][j];
               if(in[ne]!=in[i])
               {
                   dei[in[ne]]++;
                   deo[in[i]]++;
               }
            }
      ll ans=INF;
      for(int i=1;i<=bc;i++)
      {
          if(!dei[i])
            ans=min(ans,(ll)nu[i]);
          else if(!deo[i])
            ans=min(ans,(ll)nu[i]);

      }
      printf("Case %d: %I64d\n",++kcas,n*n-n-ans*(n-ans)-m);
   }
    return 0;
}