【Gym-100712 #H】Bridges

题目链接

Bridges

题目描述

An edge in an undirected graph is a bridgeif after removing it the graph will be disconnected.
Given an undirected connected graph, you are allowed to add one edge between any pair of nodes so that the total number of bridges in the graph is minimized.

输入格式

The first line of input contains (T) ((1 le T le 64))that represents the number of test cases.
The first line of each test case contains two integers: (N) ((3 le N le 100,000)) and (M) ((N-1 le M le 100,000)),where (N) is the number of nodes, and (M) is the number of edges.
Each of the following (M) lines contains two space-separated integers: (X Y) ((1 le X,Y le N))((X eq Y)), which means that there is an edge between node (X) and node (Y).
It is guaranteed that each pair of nodes is connected by at most one edge.
Test cases are separated by a blank line.

输出格式

For each test case, print a single line with the minimum possible number of bridges after adding one edge.

样例输入

2
7 7
1 2
2 3
3 1
3 4
4 5
4 6
6 7

3 3
1 2
2 3
3 1

样例输出

1
0

题解

题意：有一个无向图，要求连接两个点后剩下的桥（割边）数量最小，输出剩下的桥数量。
很明显，如果一些点在一个强连通分量里，那么你可以把这些点当成一个点。
那么我们直接缩点，题目就变成了缩点后的树上连一条边使桥最小，那么肯定连直径了。
这里要注意一下无向图的缩点和有向图的缩点有一点点区别，无向图可以节省一些步骤，但是有向图包括无向图，可以直接用板子。
如果不会缩点的可以到这里学习一下。
上代码：

#include<bits/stdc++.h>
using namespace std;
int t,n,m;
int x,y;
struct aa{
    int to,nxt;
}p[200009];
int h[100009],len;
aa p2[200009];
int h2[200009],len2;
void add(int x,int y){
    p[++len].to=y;
    p[len].nxt=h[x];
    h[x]=len;
}
int to[200009];
int dfn[100009],low[100009];
int uu[100009],lu;
int ut;
int ans,as,ss;
void tarjan(int u,int fa){
    uu[++lu]=u;
    dfn[u]=low[u]=++ut;
    for(int j=h[u];j;j=p[j].nxt){
        if(p[j].to==fa) continue;
        if(dfn[p[j].to]==0){
            tarjan(p[j].to,u);
            low[u]=min(low[u],low[p[j].to]);
        }else low[u]=min(low[u],dfn[p[j].to]);
    }
    if(dfn[u]==low[u]){
        while(uu[lu]!=u) to[uu[lu--]]=u;
        to[uu[lu--]]=u;
    }
}
void dfs(int u,int s,int fa){
    if(s>ans){ans=s;as=u;}
    for(int j=h2[u];j;j=p2[j].nxt){
        if(p2[j].to==fa) continue;
        dfs(p2[j].to,s+1,u);
    }
}
int main(){
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        len=len2=0;
        memset(h,0,sizeof(h));
        memset(h2,0,sizeof(h2));
        memset(dfn,0,sizeof(dfn));
        memset(low,0,sizeof(low));
        for(int j=1;j<=m;j++){
            scanf("%d%d",&x,&y);
            add(x,y);
            add(y,x);
        }
        tarjan(1,0);
        ss=0;
        for(int u=1;u<=n;u++){
            for(int j=h[u];j;j=p[j].nxt){
                x=to[u];y=to[p[j].to];
                if(x!=y){
                    p2[++len2].to=y;
                    p2[len2].nxt=h2[x];
                    h2[x]=len2;
                    ss++;
                }
            }
        }
        ss/=2;
        ans=0;
        dfs(to[1],0,0);
        dfs(as,0,0);
        printf("%d
",ss-ans);
    }
    return 0;
}