HDOJ 5305 Friends 暴力枚举

暴力枚举边的状态.....

Friends

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 120    Accepted Submission(s): 39

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

23 31 22 33 14 41 22 33 44 1

 

Sample Output

02

 

Source

2015 Multi-University Training Contest 2

 

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/* ***********************************************Author        :CKbossCreated Time  :2015年07月23日 星期四 12时45分03秒File Name     :1006.cpp************************************************ */#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <cmath>#include <cstdlib>#include <vector>#include <queue>#include <set>#include <map>using namespace std;vector<int> G[10];/// 1      0int on[10],off[10];int du[10];int n,m;int color[10][10];int ans;void init(){    for(int i=0;i<=n;i++)     {        G[i].clear();         on[i]=off[i]=0;        du[i]=0;    }    ans=0;    memset(color,-1,sizeof(color));}void dfs(int u){    if(u==n+1)     {        int fg=1;        for(int i=1;i<=n&&fg;i++) { if(on[i]!=off[i]) fg=0; }        ans+=fg;        return ;    }    vector<int> vi;    for(int i=0;i<du[u];i++)    {        int v=G[u][i];        if(color[u][v]!=-1) continue;        vi.push_back(v);    }    int nt=vi.size();    for(int i=0;i<(1<<nt);i++)    {        for(int j=0;j<nt;j++)        {            int v=vi[j];            int cc=(i&(1<<j));            if(cc) cc=1;            color[u][v]=cc;            color[v][u]=cc;            if(cc==0)            {                off[u]++; off[v]++;            }            else            {                on[u]++; on[v]++;            }        }        if(on[u]==off[u]) dfs(u+1);        /// uncolor        for(int j=0;j<nt;j++)        {            int v=vi[j];            int cc=(i&(1<<j));            if(cc) cc=1;            color[u][v]=-1;            color[v][u]=-1;            if(cc==0)            {                off[u]--; off[v]--;            }            else            {                on[u]--; on[v]--;            }        }    }}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    int T_T;    scanf("%d",&T_T);    while(T_T--)    {        scanf("%d%d",&n,&m);        init();        for(int i=0,u,v;i<m;i++)        {            scanf("%d%d",&u,&v);            G[u].push_back(v); G[v].push_back(u);        }        bool flag=true;        for(int i=1;i<=n&&flag;i++)        {            du[i]=G[i].size();            if(du[i]%2) flag=false;        }        if(flag)        {            dfs(1);            printf("%d\n",ans);        }        else        {            printf("%d\n",0);        }    }        return 0;}