Jimmy’s Assignment

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)

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

这题真的很坑！！937ms过的，期间有的代码好像搞错了，竟然速度只400+ms.其实答案就是n/2。

主要是bool数组真的很省时间！！

#include"stdio.h"
#include"string.h"
#include"queue"
#include"vector"
using namespace std;
#define N 5005
#define M 30005
int lx[N],ly[N];
bool mark[N];
vector<int>g[N];
int find(int k)
{
    int i,v;
    for(i=0;i<g[k].size();i++)
    {
        v=g[k][i];
        if(!mark[v])
        {
            mark[v]=1;
            if(ly[v]==-1||find(ly[v]))
            {
                ly[v]=k;lx[k]=v;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int i,u,v,n,T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(i=0;i<=n;i++)
            g[i].clear();
        for(i=0;i<n*3/2;i++)
        {
            scanf("%d%d",&u,&v);
            g[u].push_back(v);
            g[v].push_back(u);
        }
        memset(lx,-1,sizeof(lx));
        memset(ly,-1,sizeof(ly));
        int ans=0;
        for(i=1;i<=n;i++)
        {
            if(lx[i]!=-1)
                continue;
            //memset(mark,0,(n+2)*sizeof(int));
            memset(mark,0,sizeof(mark));
            ans+=find(i);
        }
        printf("%d\n",ans/2);
    }
    return 0;
}

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