ZJUT1321-Dividing //MAXMATCH

Dividing 
Time Limit:1000MS  Memory Limit:32768K

Description:

There are N(N is an even number) people in ZJUT. each person in ZJUT, loves some ones and hates the others. Is it possible to divide them to N/2 groups in which the two people love each other?

 

Input:

The first line gives a number T, means the number of test cases. The first line of each test case gives a number N, and then follows a N*N matrix. The elements of the matrix are either 0 or 1. If the j-th number in the i-th row is 1, it means the i-th person loves the j-th person, else he doesn’t love him. N,T <= 16

Output:

If it is possible to divide them, print “Yes”, else print “No”.

Sample Input:

2
4
1 1 1 0
1 1 0 1
1 1 1 0
0 1 0 1
4
1 1 0 0
0 1 0 1
1 0 1 0
0 0 1 1

Sample Output:

Yes
No

Source:

lily

这是我见过的最慢的OJ。。。

构图注意匹配的条件，要互相喜欢，图不是对称的

#include<stdio.h>

#include<string.h>

int n;

bool mat[20][20],used[20];

int link[20];

bool can(int t)

{

    for(int i=1;i<=n;i++)

    if(used[i]==false&&mat[t][i]&&mat[i][t])

    {

        used[i]=true;

        if(link[i]==-1||can(link[i]))

        {

            link[i]=t;

            return true;

        }

    }

    return false;

}

int MaxMatch()

{

    int num=0;

    memset(link,-1,sizeof(link));

    for(int i=1;i<=n;i++)

    {

        memset(used,false,sizeof(used));

        if(can(i)) num++;

    }

    return num;

}

int main()

{

    int T;

    scanf("%d",&T);

    while(T--)

    {

        scanf("%d",&n);

        for(int i=1;i<=n;i++)

          for(int j=1;j<=n;j++)

          {

              int x;

              scanf("%d",&x);

              if(i!=j) mat[i][j]=x; else mat[i][j]=0;

          }

        if(MaxMatch()==n) printf("Yes/n");

        else printf("No/n");

    }

    return 0;

}