G

题意：给你一个图，问最多能添加多少条边使图仍为不是强连通图，如果原图是强连通输出 ‘-1’

分析：先把求出连通分量进行缩点，因为是求最多的添加边，所以可以看成两部分 x，y，只能一部分向另外一部分连边，内部的就是完全图，所以是x*（x+1）+x*y+y*（y+1）-M，只需要求出来出度或者入度为0的最少点的那个连通分量即可。

**********************************************************************

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

const int MAXN = 1e5+5;
const int oo = 1e9;

struct Edge{int v, next;}e[MAXN];
int Head[MAXN], cnt;
void AddEdge(int u, int v)
{
    e[cnt].v = v;
    e[cnt].next = Head[u];
    Head[u] = cnt++;
}

int dfn[MAXN], low[MAXN], Index;
int Stack[MAXN], top, inStack[MAXN];
int blg[MAXN], bnt, nblg[MAXN];///属于哪个连通分量，连通分量里面有几个点
int outEdge[MAXN], inEdge[MAXN];

void InIt(int N)
{
    cnt = Index = top = bnt = 0;
    for(int i=0; i<=N; i++)
    {
        Head[i] = -1;
        dfn[i] = 0;
        nblg[i] = 0;
        outEdge[i] = 0;
        inEdge[i] = 0;
    }
}
void Tarjan(int u)
{
    int v;

    low[u] = dfn[u] = ++Index;
    Stack[++top] = u;
    inStack[u] = true;

    for(int j=Head[u]; j!=-1; j=e[j].next)
    {
        v = e[j].v;
        if( !dfn[v] )
        {
            Tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if(inStack[v] == true)
            low[u] = min(low[u], dfn[v]);
    }

    if(low[u] == dfn[u])
    {
        ++bnt;
        do
        {
            v = Stack[top--];
            inStack[v] = false;
            blg[v] = bnt;
            nblg[bnt]++;
        }
        while(u != v);
    }
}

int main()
{
    int T, t=1;

    scanf("%d", &T);

    while(T--)
    {
        int i, j, u, v, N, M;

        scanf("%d%d", &N, &M);

        InIt(N);

        for(i=0; i<M; i++)
        {
            scanf("%d%d", &u, &v);
            AddEdge(u, v);
        }

        for(i=1; i<=N; i++)
        {
            if( !dfn[i] )
                Tarjan(i);
        }

        for(i=1; i<=N; i++)
        for(j=Head[i]; j!=-1; j=e[j].next)
        {
            v = e[j].v;
            if(blg[i] != blg[v])
            {
                inEdge[ blg[v] ]++;
                outEdge[ blg[i] ]++;
            }
        }

        int x, y=oo;

        for(i=1; i<=bnt; i++)
        {
            if(!outEdge[i] || !inEdge[i])
                y = min(y, nblg[i]);
        }

        x = N-y;

        if(bnt == 1)
            printf("Case %d: -1
", t++);
        else
            printf("Case %d: %lld
",t++, (long long)x*(x-1)+x*y+y*(y-1)-M);
    }

    return 0; 

}