Jimmy’s Assignment（HDU 1845）

Problem Description

Jimmy is studying Advanced Graph Algorithms at his university. His most recent assignment is to find a maximum matching in a special kind of graph. This graph is undirected, has N vertices and each vertex has degree 3. Furthermore, the graph is 2-edge-connected (that is, at least 2 edges need to be removed in order to make the graph disconnected). A matching is a subset of the graph’s edges, such that no two edges in the subset have a common vertex. A maximum matching is a matching having the maximum cardinality.
  Given a series of instances of the special graph mentioned above, find the cardinality of a maximum matching for each instance.

Input

The first line of input contains an integer number T, representing the number of graph descriptions to follow. Each description contains on the first line an even integer number N (4<=N<=5000), representing the number of vertices. Each of the next 3*N/2 lines contains two integers A and B, separated by one blank, denoting that there is an edge between vertex A and vertex B. The vertices are numbered from 1 to N. No edge may appear twice in the input.

Output

For each of the T graphs, in the order given in the input, print one line containing the cardinality of a maximum matching.

Sample Input

2
4
1 2
1 3
1 4
2 3
2 4
3 4
4
1 2
1 3
1 4
2 3
2 4
3 4

Sample Output

2
2

题意：给出一 3-正则图，如果想让该图不连通，至少需要删除两条边（每个点都处于一个环中且原图连通），现在要尽量找出多的点对，使得每对点中都有边

思路：对边的两点 x、y 做两条边，然后找最大匹配数，由于左右两侧所代表的点是相同的，比如从左边 1 出发选择了右边的 3，后面的匹配时右边的 1 也就不能选了，为解决这种问题，需要构造双向边，然后跑最大匹配，最后结果除以 2 即可

Source Program

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define PI acos(-1.0)
#define E 1e-6
#define MOD 16007
#define INF 0x3f3f3f3f
#define N 10001
#define LL long long
using namespace std;
bool vis[N];
int link[N];
vector<int> G[N];
bool dfs(int x){
    for(int i=0;i<G[x].size();i++){
        int y=G[x][i];
        if(!vis[y]){
            vis[y]=true;
            if(link[y]==-1 || dfs(link[y]))	{
                link[y]=x;
                return true;
            }
        }
    }
    return false;
}
int hungarian(int n)
{
    int ans=0;
    for(int i=1;i<=n;i++){
        memset(vis,false,sizeof(vis));
        if(dfs(i))
            ans++;
    }
    return ans;
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);

        memset(link,-1,sizeof(link));
        for(int i=0;i<N;i++)
            G[i].clear();

        int x,y;
        for(int i=0;i<n*3/2;i++){
            scanf("%d%d",&x,&y);
            G[x].push_back(y);
            G[y].push_back(x);
        }

        printf("%d\n",hungarian(n)/2);

    }
    return 0;
}

————————————————————————————————————————————————————

大神的思路：点击这里

对于一 n 个点的三正则图，求最大匹配。

根据握手定理，n 一定是偶数。由于三正则图，而且题目提示是 2 边连通，所以图中不存在桥，也就是一定可以找到一条回路经过每个顶点至少一次（强连通的定义：强连通图一定存在一条回路记过每个顶点至少一次）

由于是三则图，每个顶点的度是 3，如果这条回路经过某个顶点 2 次，那么这个顶点的度就是 4，与条件矛盾。

因此，这条经过每个顶点一次的交错路就可以作出 n/2 匹配。

#include<stdio.h>
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int n;
        scanf("%d",&n);
        int x,y;
        for(int i=1;i<=n*3/2;i++)
            scanf("%d%d",&x,&y);
        printf("%d\n",n/2);
    }
    return 0;
}