Strongly connected HDU

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.
A strongly connected digraph is a directed graph in which it is possible to reach any node starting from any other node by traversing edges in the direction(s) in which they point.

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

33 31 22 33 13 31 22 31 36 61 22 33 14 55 66 4

Sample Output

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

题意：给一个简单无向图，让你加最多的边，使他还是一个简单无向图。

思路：首先要把这个图变成一个完全图，然后减去最初的m条边。然后想要加入的边最大化，你需要强联通缩点，把入度为0或者出度为0的内含节点最少的联通块找出来，然后再减去最小联通块内的点与其他点的连接边就可以了。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define ll __int64const int N = 100005;struct EDG{    int to,next;} edg[N];int eid,head[N];int low[N],dfn[N],vist[N],num[N],id[N],deep,stack1[N],tn,top;int in[N],out[N];void init(){    eid=tn=top=deep=0;    memset(head,-1,sizeof(head));    memset(vist,0,sizeof(vist));    memset(in,0,sizeof(in));    memset(out,0,sizeof(out));    memset(num,0,sizeof(num));}void addEdg(int u,int v){    edg[eid].to=v;    edg[eid].next=head[u];    head[u]=eid++;}void tarjer(int u){    stack1[++top]=u;    vist[u]=1;    deep++;    low[u]=dfn[u]=deep;    for(int i=head[u]; i!=-1; i=edg[i].next)    {        int v=edg[i].to;        if(vist[v]==0)        {            vist[v]=1;            tarjer(v);            low[u]=min(low[u],low[v]);        }        else if(vist[v]==1)            low[u]=min(low[u],dfn[v]);    }    if(low[u]==dfn[u])    {        tn++;        do        {            vist[stack1[top]]=2;            num[tn]++;            id[stack1[top]]=tn;        }        while(stack1[top--]!=u);    }}ll solve(int n,int m){    ll ans=n*(n-1)-m;    int minnum=N;    for(int i=1; i<=n; i++)        if(vist[i]==0)            tarjer(i);    if(tn==1) return -1;    for(int u=1; u<=n; u++)        for(int i=head[u]; i!=-1; i=edg[i].next)        {            int v=edg[i].to;            if(id[u]!=id[v])                in[id[v]]++,out[id[u]]++;        }    for(int i=1; i<=tn; i++)        if(in[i]==0||out[i]==0)        {            minnum=min(minnum,num[i]);        }    ans-=minnum*(n-minnum);    return ans;}int main(){    int T,n,m,c=0,a,b;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&m);        init();        for(int i=1; i<=m; i++)        {            scanf("%d%d",&a,&b);            addEdg(a,b);        }        printf("Case %d: %I64d\n",++c,solve(n,m));    }}