zoj 3605 Find the Marble

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4708

这题的dp还是不难想的，然后把状态转移时候的概率算上去就可以了

代码中有注释

#include<iostream> 
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
#define N 55
int a[N],b[N];
double c[N][N];  //c[i][j] = i个里面取j个的取法
double dp[N][N][N];  //dp[i][j][l] 前i次操作看了j次在l位瓶子里面的概率
int main(){
    for(int i=0;i<=50;++i){
        c[i][0]=1;
        for(int j=1;j<=i;++j) c[i][j]=c[i-1][j-1]+c[i-1][j];
    }
    int T,m,n,k,s,t;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d%d%d",&n,&m,&k,&s);
        for(int i=1;i<=m;++i) scanf("%d%d",&a[i],&b[i]);
        memset(dp,0,sizeof(dp));
        dp[0][0][s]=1;
        for(int i=0;i<m;++i) for(int j=0;j<=i;++j) for(int l=1;l<=n;++l){
            dp[i+1][j][l]+=dp[i][j][l]*c[m-i-1][k-j]/c[m-i][k-j];
            if(a[i+1]==l) t=b[i+1];
            else if(b[i+1]==l) t=a[i+1];
            else t=l;
            dp[i+1][j+1][t]+=dp[i][j][l]*c[m-i-1][k-j-1]/c[m-i][k-j];
        }
        int ans=1;
        for(int i=2;i<=n;++i)
            if(dp[m][k][i]>dp[m][k][ans]) ans=i;
        printf("%d\n",ans);
    }
    return 0;
}