2016 多校1 1005 二分图最大匹配

Problem Description
SJX has 2*N magic gems. N of them have Yin energy inside while others have Yang energy. SJX wants to make a necklace with these magic gems for his beloved BHB. To avoid making the necklace too Yin or too Yang, he must place these magic gems Yin after Yang and Yang after Yin, which means two adjacent gems must have different kind of energy. But he finds that some gems with Yang energy will become somber adjacent with some of the Yin gems and impact the value of the neckless. After trying multiple times, he finds out M rules of the gems. He wants to have a most valuable neckless which means the somber gems must be as less as possible. So he wonders how many gems with Yang energy will become somber if he make the necklace in the best way.

Input
Multiple test cases.

For each test case, the first line contains two integers N(0≤N≤9),M(0≤M≤N∗N), descripted as above.

Then M lines followed, every line contains two integers X,Y, indicates that magic gem X with Yang energy will become somber adjacent with the magic gem Y with Yin energy.

Output
One line per case, an integer indicates that how many gem will become somber at least.

Sample Input
2 1
1 1
3 4
1 1
1 2
1 3
2 1

Sample Output
1
1

8！暴力枚举阴的部分建图后，转化为二分图最大匹配问题。匈牙利算法跑一遍即可。
之所以是8！而不是9！是因为环的对称性，可以直接set第一个是1。

代码：

#include <bits/stdc++.h>
using namespace std;
int a[15];
int n, m;
int Max;
bool vis[15];
bool vis2[15];
int linker[15];
bool mp[15][15];
vector<int>g[15];
struct match{
    int x, y;
};
//map <match, int>mp;
bool operator<(match const&a, match const&b)
{
    if(a.x!=b.x)return a.x<b.x;
    else return a.y<b.y;
}

bool dfs2(int u)
{
    for(int v=0;v<g[u].size();v++)
    {
        int tmp=g[u][v];
        if(!vis2[tmp])
        {
            vis2[tmp]=true;
            if(linker[tmp]==-1||dfs2(linker[tmp]))
            {
                linker[tmp]=u;
                return true;
            }
        }
    }
    return false;
}

int hungary()
{
    for(int i=1;i<=n;i++)g[i].clear();
    int x1, x2, y1, y2;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            x1=i;x2=i;
            y1=a[j];
            if(j==n)y2=a[1];
            else y2=a[j+1];
            if(!mp[x1][y1]&&!mp[x2][y2])g[i].push_back(j);
        }
    }
    //printf("pic is:\n");
    /*for(int i=1;i<=n;i++)
    {
        for(int j=0;j<g[i].size();j++)
            printf("%d ", g[i][j]);
        printf("\n");
    }*/
    int res=0;
    memset(linker, -1, sizeof(linker));
    for(int u=1;u<=n;u++)
    {
        memset(vis2, false, sizeof(vis2));
        if(dfs2(u))res++;
    }
    return res;
}

void dfs(int flor)
{
    if(flor==n+1)
    {
        int res=hungary();
        Max=max(Max, res);
        //for(int i=1;i<=n;i++)
        //    printf("%d%c", a[i],i==n?'\n':' ');
        return;
    }
    for(int i=2;i<=n;i++)
    {
        if(!vis[i])
        {
            vis[i]=true;
            a[flor]=i;
            dfs(flor+1);
            vis[i]=false;
        }
    }
    return;
}
int main()
{
    while(~scanf("%d%d", &n, &m))
    {
        //mp.clear();
        memset(mp, 0, sizeof(mp));
        int x, y;
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d", &x, &y);
            //match tmp;
            //tmp.x=x;tmp.y=y;
            mp[x][y]=true;
        }
        Max=0;
        memset(vis, false, sizeof(vis));
        a[1]=1;
        dfs(2);
        printf("%d\n",n-Max);
    }
    return 0;
}