Friends

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

There are n people
and m pairs
of friends. For every pair of friends, they can choose to become online friends (communicating using online applications) or offline friends (mostly using face-to-face communication). However, everyone in these n people
wants to have the same number of online and offline friends (i.e. If one person has x onine
friends, he or she must have x offline
friends too, but different people can have different number of online or offline friends). Please determine how many ways there are to satisfy their requirements.

Input

The first line of the input is a single integer T (T=100),
indicating the number of testcases.

For each testcase, the first line contains two integers n (1≤n≤8) and m (0≤m≤n(n?1)2),
indicating the number of people and the number of pairs of friends, respectively. Each of the next m lines
contains two numbers x and y,
which mean x and y are
friends. It is guaranteed that x≠y and
every friend relationship will appear at most once.

Output

For each testcase, print one number indicating the answer.

Sample Input

2
3 3
1 2
2 3
3 1
4 4
1 2
2 3
3 4
4 1

Sample Output

0
2

Source

/*

题意：n个人，m个关系，每个人和别人的关喜有线上和线下，求每个人线上和线下的关系一样多的方案数
思路: dfs
先贴别人代码，跑的快
，我的跑的慢

*/

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
using namespace std;

typedef __int64 LL;

#define N 100

int x[N],y[N],in[N],on[N],off[N];
int n,m;
int ans;

bool judge()
{
    if(m&1) return false;
    for(int i=1;i<=n;i++) if(in[i]&1) return false;
    return true;
}

void dfs(int pos)
{
    if(pos==m)
    {
        ans++;
        return ;
    }
    int u=x[pos],v=y[pos];
    if(on[u]<in[u]/2&&on[v]<in[v]/2)  //这个边为online 边
    {
        on[u]++;
        on[v]++;
        dfs(pos+1);
        on[u]--;
        on[v]--;
    }

    if(off[u]<in[u]/2&&off[v]<in[v]/2) //这个边为off边
    {
        off[u]++;
        off[v]++;
        dfs(pos+1);
        off[u]--;
        off[v]--;
    }
}
int main()
{
    int i,j,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        memset(in,0,sizeof(in));
        memset(on,0,sizeof(on));
        memset(off,0,sizeof(off));

        for(i=0;i<m;i++)
        {
            scanf("%d%d",&x[i],&y[i]);
            in[x[i]]++;
            in[y[i]]++;
        }
        ans=0;
        if(!judge())
        {
            printf("0\n");
            continue;
        }
        dfs(0);
        printf("%d\n",ans);
    }
    return 0;
}

/*
我的代码
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define bug printf("hihi\n")

#define eps 1e-8
typedef __int64 ll;

using namespace std;

#define INF 0x3f3f3f3f

#define N 1<<8

__int64 ans;

int on[N],down[N];
int n,m,f[N],in[N];

inline int get(int x)
{
    int s=0;
    while(x)
    {
        s++;
        x&=(x-1);
    }
    return s;
}

inline bool judge(int pos,int cur)
{
    int i;
    for(i=0;i<n;i++)
       if(on[i]&(1<<pos)&&!(cur&(1<<i))) return false;

    for(i=0;i<n;i++)
      if(down[i]&(1<<pos)&&(cur&(1<<i))) return false;
    return true;
}

void dfs(int pos)
{
    if(pos==n)
    {
        ans++;
        return ;
    }
    int i,len=1<<n;
    for(i=0;i<len;i++)
    {
        if((f[pos]&i)!=i) continue;
        int tt=get(i);
        if(tt!=in[pos]/2) continue;
        if(!judge(pos,i)) continue;
        on[pos]=i;
        down[pos]=f[pos]^i;
        dfs(pos+1);
        on[pos]=0;
        down[pos]=0;
    }
}

int main()
{
    int i,j,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        int u,v;
        memset(f,0,sizeof(f));
        memset(on,0,sizeof(on));
        memset(down,0,sizeof(down));
        memset(in,0,sizeof(in));
        bool flag=false;
        if(m&1) flag=true;
        while(m--)
        {
            scanf("%d%d",&u,&v);
            u--;v--;
            f[u]|=1<<v;
            f[v]|=1<<u;
            in[u]++;
            in[v]++;
        }

        for(i=0;i<n;i++)
        {
            if(in[i]&1) flag=true;
        }
        if(flag)
        {
            printf("0\n");
            continue;
        }
        ans=0;
        int len=1<<n;
        for(i=0;i<len;i++)
        {
            if((f[0]&i)!=i) continue;
            int tt=get(i);
            if(tt!=in[0]/2) continue;
            on[0]=i;
            down[0]=f[0]^i;
            dfs(1);
            on[0]=0;
            down[0]=0;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。