POJ 3352 Road Construction&& POJ 3177 Redundant Paths 双联通分量

大意：给定n点，和m条边的关系图中的一些边随时可能施工导致不能够通过，所以至少加多少条边才能够使得途中任意两条边联通？

思路：很明显只要图中的任意两点都是两条边来链接即可。那么我们可以先缩点构建新图，然后统计出度为1的点的个数ans，那么需要加的边数就是（ans+1）/2条;

(PS;因为建图是双向的图所以，在Tarjan缩点的时候就需要遇到临边便越过，并且判断是不是同一个联通分支用num比较！)

#include<map>
#include<queue>
#include<cmath>
#include<cstdio>
#include<stack>
#include<iostream>
#include<cstring>
#include<algorithm>
#define LL int
#define inf 0x3f3f3f3f
#define eps 1e-8
#include<vector>
#define ls l,mid,rt<<1
#define rs mid+1,r,rt<<1|1
using namespace std;
int n,m;
const int Ma = 1000005;

int head[Ma],dfn[Ma],num[Ma],du[Ma],stk[Ma],vis[Ma],low[Ma];
int cnt,top,time,css;
bool mp[5100][5100];
struct node{
    int to,next;
}q[Ma];

void bu(int a,int b){
    q[cnt].to = b;
    q[cnt].next = head[a];
    head[a] = cnt++;
}

void init(){
    time = 1;
    css = top = cnt = 0;
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(vis,0,sizeof(vis));
    memset(head,-1,sizeof(head));
    memset(num,0,sizeof(num));
    memset(du,0,sizeof(du));
    memset(mp,false,sizeof(mp));
}

void Tarjan(int u,int To){
    low[u] = dfn[u] = time++;
    vis[u] = 1;
    stk[top++] = u;
    for(int i = head[u]; ~i ; i = q[i].next){
        int v = q[i].to;
        if(i == (To^1)) continue;
        if(!vis[v]){
            Tarjan(v,i);
            low[u] = min(low[u],low[v]);
        }
        else
            low[u] = min(low[u],dfn[v]);

    }
    if(low[u] == dfn[u]){//找到极大联通分量
        css++;
        while(top > 0&&stk[top] != u){
            top --;
            vis[stk[top] ] = 2;
            num[stk[top] ] = css;
        }
    }

}

int main(){
    int k,i,j,a,b;
    while(~scanf("%d%d",&n,&m)){
        init();
        for(i = 0;i < m;++ i){
            scanf("%d%d",&a,&b);//////注意重边的问题！
            if(mp[a][b]){
                continue;
            }
            mp[a][b] = mp[b][a] = true;
            bu(a,b);
            bu(b,a);
        }
        Tarjan(1,-1);
        for(i = 1;i <= n;++ i)
            for(j = head[i]; ~j ;j = q[j].next)
                if( num[i] != num[q[j].to] )
                    du[num[i] ]++;
        int ans = 0;
        for(i = 1;i <= n;++ i)
            if(du[i] == 1)
                ans++;
        printf("%d\n",(ans+1)>>1);
    }
    return 0;
}