poj 2151 Check the difficulty of problems 概率dp

把存在轉化爲所有 p=1-!p

#include<iostream>
#include<algorithm>
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<queue>
#include<cmath>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
int dir[4][2]={0,1,0,-1,1,0,-1,0};
const double eps=1e-8;

int m,t,n;
double p[1005][33];
double dp[1005][33][33];

int main()
{
	while(scanf("%d%d%d",&m,&t,&n)!=EOF&&m!=0)
	{
		int i,j,k;
		for(i=1;i<=t;i++)
			for(j=1;j<=m;j++)
				scanf("%lf",&p[i][j]);
		double sum1=1,sum2=1,s=0;
		for(k=1;k<=t;k++)
		{
			dp[k][1][1]=p[k][1];
			dp[k][1][0]=1-p[k][1];
			for(i=2;i<=m;i++)
			{
				dp[k][i][0]=dp[k][i-1][0]*(1-p[k][i]);
				for(j=1;j<i;j++)
					dp[k][i][j]=dp[k][i-1][j]*(1-p[k][i])+dp[k][i-1][j-1]*p[k][i];
				dp[k][i][i]=dp[k][i-1][i-1]*p[k][i];
			}
			sum1*=(1-dp[k][m][0]);
			s=0;
			for(i=1;i<n;i++)
				s+=dp[k][m][i];
			sum2*=s;
		}
		printf("%.3lf\n",sum1-sum2);
	}
	return 0;
}