hdu5727 Necklace（二分图）

思路：枚举阴球的全排列，然后如果和阳球不冲突的话就在这个位置和阳球连一条边，求二分图最大匹配，最后n-最大匹配就是冲突的最小个数

#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
const int maxn=120+5;
#define INF 999999
int ans = 0;
int vis[maxn],pos[maxn];
int n,m;
int mp[20][20];
struct Max_Match
{
    int n;
    vector<int> g[maxn];
//	int g[maxn][maxn];
    bool vis[maxn];
    int left[maxn];

    void init(int n)
    {
        this->n=n;
        for(int i=1; i<=n; ++i) g[i].clear();
	   // memset(g,0,sizeof(g));
        memset(left,-1,sizeof(left));
    }

    bool match(int u)
    {
        for(int i=0;i<g[u].size();++i)
        {
            int v=g[u][i];
            if(!vis[v])
            {
                vis[v]=true;
                if(left[v]==-1 || match(left[v]))
                {
                    left[v]=u;
                    return true;
                }
            }
        }
        return false;
    }

    int solve()
    {
        int ans=0;
        for(int i=1; i<=n; ++i)
        {
            memset(vis,0,sizeof(vis));
            if(match(i)) ++ans;
        }
        return ans;
    }
}MM;
void work()
{
     MM.init(n*2);
	 for(int i = 2;i<=n;i++)
		 for(int j = 1;j<=n;j++)
		 {
			  if(mp[pos[i]][j] || mp[pos[i-1]][j])
				  continue;
			  MM.g[i].push_back(j);
		 }
	 for(int i = 1;i<=n;i++)
	 {  
		 if(mp[pos[1]][i] || mp[pos[n]][i])
			 continue;
         MM.g[1].push_back(i);
	 }
	 ans = min(ans,n-MM.solve());
}
int jianzhi = 0;
void shengchengquanpailie(int cnt)
{
//	 jianzhi++;
//	 if(jianzhi>100000)
//		 return;
	 if(!ans)
		 return;
     if(cnt==n+1)
	 {
		 work();
		 return;
	 }
     for(int i = 1;i<=n;i++)
	 {
		 if(vis[i])
			 continue;
		 pos[cnt]=i;
		 vis[i]=1; 
		 shengchengquanpailie(cnt+1);
		 pos[cnt]=0;
		 vis[i]=0;
	 }
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
		if(n==0)
		{
			printf("0\n");
			continue;
		}
		memset(vis,0,sizeof(vis));
		memset(pos,0,sizeof(pos));
		memset(mp,0,sizeof(mp));
		jianzhi = 0;
        while(m--)
        {
            int u,v;
            scanf("%d%d",&u,&v);
			mp[v][u]=1;
        }
        ans = INF;
		pos[1]=1;
		vis[1]=1;
		shengchengquanpailie(2);
		printf("%d\n",ans);
    }
    return 0;
}

Problem Description

N

 

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  Ywith 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

 

Author

HIT

 

Source

2016 Multi-University Training Contest 1