Arbiter

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1242 Accepted Submission(s): 609

Problem Description

Arbiter is a kind of starship in the StarCraft science-fiction series. The Arbiter-class starship is a Protoss warship specializing in providing psychic support. Arbiters were crewed exclusively by Judicators; unlike other warships that were manned predominantly by Templar. The Judicator used the Arbiter as a base to provide support using space-time manipulation.
Arbiters could weaken space-time, tearing rifts in the fabric of space-time, creating a vortex linking another location to the Arbiter’s location. This could be used to move personnel over long distances between stars.
In the meantime of widely used Arbiter to transfer, KMXS, the captain of one Arbiter, was warning that some person had got a serious mental disorder after the trip on his Arbiter. By using mice as model animals, he found the sake, it’s because of chirality!
Every person has chirality, either left-handed or right-handed. Actually all the persons must live with the food which has the same chirality. When one person took Arbiter from one star to another one, his chirality will be changed (from left-handed to right-handed or from right-handed to left-handed). If a person took a long trip and finally got back to his own star, however, his chirality might be changed to the opposite state other than his original, which would cause fatal mental disorder, or even death.
KMXS has the channels map among the starts and he need to prohibit minimum number of channels from traveling so that wherever a person starts his traveling from when he gets his original star he’ll be safe. KMXS turns to your help.

Input

The first line of input consists of an integer T, indicating the number of test cases.
The first line of each case consists of two integers N and M, indicating the number of stars and the number of channels. Each of the next M lines indicates one channel (u, v) which means there is a bidirectional channel between star u and star v (u is not equal to v).

Output

Output one integer on a single line for each case, indicating the minimum number of channels KMXS must prohibit to avoid mental disorder.

Constraints
0 < T <= 10
0 <= N <= 15 0 <= M <= 300
0 <= u, v < N and there may be more than one channel between two stars.

Sample Input

   1 3 3 0 1 1 2 2 0  

Sample Output

1

Source

2009 Asia Wuhan Regional Contest Online

题目大意：

给你一张有向图，问你最少删除的边数使得图中没有奇数环，即环的点数为奇数（1除外）

题目思路：

首先我们看看二分图的定义：无向图G=<V,E>为二分图的充要条件是G的所有回路的长度均为偶数。

所以我们可以利用二分图的性质来做这题，首先我们可以利用二进制的来枚举他的所有二分图的组合，二进制中1和0分别属于不同的集合，然后我们在枚举同一集合当中的点是否存在边，如果存在边就删掉该边，然后在记录删除的次数，最后取个最小值就是我们要求的答案

AC代码：

#include<cstring>
#include<cstdio>
#define min(x,y) (x<y?x:y)
int mp[20][20];
int n,m;

int sove(int k){
    int num=0;
    for(int i=0;i<n;i++){
        for(int j=i+1;j<n;j++){
            if((1&(k>>i))==(1&(k>>j))&&mp[i][j])num+=mp[i][j];    //枚举所有同一集合的点的组合看是否有边
        }
    }
    return num;
}

int main()
{
    int t;scanf("%d",&t);
    while(t--){
        memset(mp,0,sizeof(mp));
        scanf("%d%d",&n,&m);
        while(m--){
            int u,v;scanf("%d%d",&u,&v);
            mp[u][v]++,mp[v][u]++;       //记录边出现的次数
        }
        int ans = 1e8;
        for(int i=1;i<(1<<n);i++){   //枚举所有状态
            ans=min(ans,sove(i));   //取所有状态的最小值
        }
        if(ans==1e8)ans=0;
        printf("%d\n",ans);
    }
    return 0;
}

HDU3118-利用二进制状态和二分图的性质（好题）

Arbiter