題目連結

Bridges

題目描述

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.

輸入格式

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: \(X\ Y\) (\(1 \le X,Y \le N\))(\(X \neq Y\)), 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.

輸出格式

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

樣例輸入

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

樣例輸出

1
0

題解

題意：有一個無向圖，要求連線兩個點後剩下的橋（割邊）數量最小，輸出剩下的橋數量。
很明顯，如果一些點在一個強連通分量裡，那麼你可以把這些點當成一個點。
那麼我們直接縮點，題目就變成了縮點後的樹上連一條邊使橋最小，那麼肯定連直徑了。
這裡要注意一下無向圖的縮點和有向圖的縮點有一點點區別，無向圖可以節省一些步驟，但是有向圖包括無向圖，可以直接用板子。
如果不會縮點的可以到這裡學習一下。
上程式碼：

#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\n",ss-ans);
    }
    return 0;
}