UVA 11600-Masud Rana（状压，概率dp）

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <vector>
#include <string>
#include <cctype>
#include <complex>
#include <cassert>
#include <utility>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
typedef pair<int,int> PII;
typedef long long ll;
#define lson l,m,rt<<1
#define pi acos(-1.0)
#define rson m+1,r,rt<<11
#define All 1,N,1
#define N 50
#define read freopen("in.txt", "r", stdin)
const ll  INFll = 0x3f3f3f3f3f3f3f3fLL;
const int INF= 0x7ffffff;
const int mod =  1000000007;
vector<int>e[N];
int used[N],num[N],n,m,len;
map<int,double>dp[N];
//统计各连通分量的节点数
int dfs(int u){
    used[u]=1;
    int total=1;
    for(int i=0;i<e[u].size();++i){
        if(!used[e[u][i]])
            total+=dfs(e[u][i]);
    }
    return total;
}
//记忆化搜索
double solve(int i,int s){
    if(dp[i].count(s))return dp[i][s];
    int liantong=0;
    //当前联通的节点数
    for(int j=0;j<len;++j){
        if(s&(1<<j))
        liantong+=num[j];
    }
    if(liantong==n)return dp[i][s]=0;
    //要选择未联通的点需要的平均天数
    dp[i][s]=1.0*(n-1)/(n-liantong);
    //选择一个未连接的联通分量
    for(int j=0;j<len;++j){
        if(!(s&(1<<j)))
        dp[i][s]+=solve(j,s|(1<<j))*num[j]/(n-liantong);
    }
    return dp[i][s];
}
int main()
{
    int t,cas=0;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        int u,v;
        memset(used,0,sizeof(used));
        memset(num,0,sizeof(num));
        for(int i=1;i<=n;++i)
            e[i].clear();
        while(m--){
            scanf("%d%d",&u,&v);
            e[u].push_back(v);
            e[v].push_back(u);
        }
        len=0;
        for(int i=1;i<=n;++i){
            if(used[i])continue;
            dp[len].clear();
            num[len++]=dfs(i);
        }
        printf("Case %d: %.6lf\n", ++cas, solve(0, 1));
    }
return 0;
}