bridge

Posted 2020-10-10 山杉三

Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges.
The i-th edge (1≤i≤M) connects Vertex ai and Vertex bi.

An edge whose removal disconnects the graph is called a bridge.
Find the number of the edges that are bridges among the M edges.

Notes

• A self-loop is an edge i such that ai=bi (1≤i≤M).
• Double edges are a pair of edges i,j such that ai=aj and bi=bj (1≤i<j≤M).
• An undirected graph is said to be connected when there exists a path between every pair of vertices.

Constraints

• 2≤N≤50
• N?1≤M≤min(N(N?1)?2,50)
• 1≤ai<bi≤N
• The given graph does not contain self-loops and double edges.
• The given graph is connected.

---

Input

Input is given from Standard Input in the following format:

N M  
a1 b1  
a2 b2
:  
aM bM

Output

Print the number of the edges that are bridges among the M edges.

---

Sample Input 1

Copy

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

Sample Output 1

Copy

4

The figure below shows the given graph:

The edges shown in red are bridges. There are four of them.

---

Sample Input 2

Copy

3 3
1 2
1 3
2 3

Sample Output 2

Copy

0

It is possible that there is no bridge.

---

Sample Input 3

Copy

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

Sample Output 3

Copy

5

It is possible that every edge is a bridge.

题目意思:判断给定的图像中的不属于环的边有几条;

主要思路:1.遍历判断每一条边中的一点能否到达另一点

 

代码1:每次判断每条边时,都创建一个地图副本,把每次走过的边都消除,

比较容易理解,但是遇到多的数据的时候,会很慢;

#include<iostream>
#include<string>
#include<algorithm>
#include <map>
#include <string.h>
 
using namespace std;
 
struct S
{
    int Map[55][55];
};
 
int N,M;
int a[55],b[55],Map[55][55];
 
bool dfs(S s,int l,int r)
{
   // cout<<" -> "<<l<<" "<<r<<endl;
    int falg = false;
    for(int i = 1;i<=N;i++){
        if((l>i&&s.Map[i][l])||(l<i&&s.Map[l][i])){
            if(l>i) s.Map[i][l] = 0;
            else s.Map[l][i] = 0;
            if(i==r)
                return true;
            if(dfs(s,i,r))
                return true;
        }
    }
    return falg;
}
 
int main()
{
    cin>>N>>M;
    for(int i = 1;i <=M; i++){
        cin>>a[i]>>b[i];
        if(a[i]>b[i])
            Map[b[i]][a[i]] = 1;
        else
            Map[a[i]][b[i]] = 1;
     }
 
     int sum = 0;
    for(int i = 1;i<=M;i++){
        S s;
        memcpy(s.Map,Map,sizeof(Map));
        if(a[i]>b[i])
            s.Map[b[i]][a[i]] = 0;
        else
            s.Map[a[i]][b[i]] = 0;
        if(!dfs(s,a[i],b[i]))
            sum++;
       // cout<<a[i]<<" "<<b[i]<<" <-> "<<sum<<endl;
    }
 
    cout<<sum<<endl;
 
    return 0;
}

代码1

 

代码2;相当与上面代码的改进,用一个临时数组来储存每次经过的头

#include<iostream>
#include<string>
#include<algorithm>
#include <string.h>
#include <stdio.h>
#include <math.h>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <stdlib.h>
using namespace std;

int N,M;
int a[55],b[55],Map[55][55];
int visit[55];

bool dfs(int l,int r,int num)
{
    int j ;
    if(l==r||Map[l][r]==1)
        return true;
    visit[num] = l;
    for(int i = 1;i<=M;i++){
        for(j=0;j<num;j++){
            if(visit[j]==i)
                break;
        }
        if(j<num&&visit[j]==i) continue;
        if(Map[l][i]!=1) continue;
        if(dfs(i,r,num+1))
            return true;
    }

    return false;
}

int main(){
    cin>>N>>M;
    for(int i = 1;i<=M;i++){
        cin>>a[i]>>b[i];
        Map[a[i]][b[i]] = Map[b[i]][a[i]] = 1;
    }

    int sum = 0;
    for(int i = 1;i<=M;i++){
        memset(visit,0,sizeof(visit));
        Map[a[i]][b[i]] = 0;
        if(!dfs(a[i],b[i],0))
            sum++;
        Map[a[i]][b[i]] = 1;
    }
    cout<<sum<<endl;
    return 0;
}

代码2

 

代码3:使用了Floyd算法,来判断俩点之间的连通性;

代码3

 

代码4:储存每个结点连接的边的个数,每次都值为1的点出发,当前结点的值和相邻的点的值都减一;

当没有值为一的点时结束;过程中值为1的点的个数就是不循环的边的个数;

代码4