UVALive 6044(双连通分量的应用)

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=34902

思路：首先是双连通缩点，然后就是搜索一下，搜索时要跳过连通分量的点的个数>=2的点，最后的答案是n*(n-1)/2.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stack>
#include<vector>
using namespace std;
#define MAXN 11111
#define MAXM 444444

struct Edge{
    int v,next;
}edge[MAXM];

int n,m,NE,cnt,_count;
int head[MAXN];

void Insert(int u,int v)
{
    edge[NE].v=v;
    edge[NE].next=head[u];
    head[u]=NE++;
}

int low[MAXN],dfn[MAXN];
int color[MAXN];
bool mark[MAXN];
stack<int>S;
void Tarjan(int u,int father)
{
    int flag=0;
    low[u]=dfn[u]=++cnt;
    mark[u]=true;
    S.push(u);
    for(int i=head[u];i!=-1;i=edge[i].next){
        int v=edge[i].v;
        if(v==father&&!flag){ flag=1;continue; }
        if(dfn[v]==0){
            Tarjan(v,u);
            low[u]=min(low[u],low[v]);
        }else if(mark[v]){
            low[u]=min(low[u],dfn[v]);
        }
    }
    if(low[u]==dfn[u]){
        _count++;
        int x=S.top();
        if(x==u)S.pop();
        else {
            do{
                x=S.top();
                S.pop();
                mark[x]=false;
                color[x]=_count;
            }while(x!=u);
        }
    }
}

int ans;
void dfs(int u,int father)
{
    color[u]=1;
    _count++;
    for(int i=head[u];i!=-1;i=edge[i].next){
        int v=edge[i].v;
        if(v==father)continue;
        if(color[v])continue;
        dfs(v,u);
    }
}


int main()
{
    int _case,u,v,t=1;
    scanf("%d",&_case);
    while(_case--){
        scanf("%d%d",&n,&m);
        NE=0;
        memset(head,-1,sizeof(head));
        while(m--){
            scanf("%d%d",&u,&v);
            Insert(u,v);
            Insert(v,u);
        }
        cnt=_count=0;
        memset(dfn,0,sizeof(dfn));
        memset(color,0,sizeof(color));
        for(int i=1;i<=n;i++){
            if(dfn[i]==0)Tarjan(i,-1);
        }
        ans=0;
        for(int i=1;i<=n;i++){
            if(color[i]==0){
                _count=0;
                dfs(i,-1);
                ans+=_count*(_count-1)/2;
            }
        }
        printf("Case #%d: %d
",t++,ans);
    }
    return 0;
}