HDU 5727 Necklace (二分图匹配hungary)

Necklace

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 11 13 41 11 21 32 1

 

Sample Output

11题意：有一些阴阳球组成一个项链，一些阳球在和某些球相邻时会变暗，要求最少的变暗个数分析：难点还是在于建模，一开始想用暴搜去解决，但是没有想到一个好的剪枝策略，再自己看看题目，似乎和二分图有关，对于一个组合，我们可以这样想，如果阳球的左右都没有与之影响的阴球，那么那个位置一定可行，否则那个位置不可取，那么题目就转换成了阳球与插入位置的二分匹配了，现在就只需要确定阴球位置，就可以建图，直接套hungary算法分析一下复杂度： 8！*81 可行
#include<cstring>#include<iostream>#include<cstdio>#include<algorithm>using namespace std;int a[15];int G[15][15];int g[10][10];int uN,vN;int linker[15];bool used[15];bool dfs(int u){    for(int v=1; v<=vN; v++)    {        if(g[u][v]&&!used[v])        {            used[v]=true;            if(linker[v]==-1||dfs(linker[v]))            {                linker[v]=u;                return true;            }        }    }    return false;}int hungary(){    int res=0;    memset(linker,-1,sizeof linker);    for(int u=1; u<=uN; u++)    {        memset(used,false,sizeof used);        if(dfs(u)) res++;    }    return res;}int main(){    int n,m;    while(scanf("%d%d",&n,&m)!=EOF)    {        if(n==0)        {            printf("0\n");            continue;        }        memset(G,0,sizeof G);        for(int i=1; i<=m; i++)        {            int a,b;            scanf("%d%d",&a,&b);            G[a][b]=1;        }        uN=vN=n;        int ans=100;        for(int i=1; i<=n; i++) a[i]=i;        do        {            for(int i=1; i<=n; i++)            {                for(int j=1; j<=n; j++)                {                    g[i][j]=0;                    if(j==n)                    {                        if(!G[i][a[j]]&&!G[i][1]) g[i][j]=1;                    }                    else if(!G[i][a[j]]&&!G[i][a[j+1]])                    {                        g[i][j]=1;                    }                }            }            int num=hungary();            ans=min(ans,n-num);        }while(next_permutation(a+2,a+n+1));        printf("%d\n",ans);    }    return 0;}