poj 1644 To Bet or Not To Bet

题目链接：http://poj.org/problem?id=1644

题目思路：用dp[i][j]记录在i位置回合数为j的概率，递推即可。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<string>
#include<queue>
#include<algorithm>
#include<vector>
#include<stack>
#include<list>
#include<iostream>
#include<map>
using namespace std;
#define inf 0x3f3f3f3f
#define Max 110
int max(int a,int b)
{
	return a>b?a:b;
}
int min(int a,int b)
{
	return a<b?a:b;
}
double dp[60][60];
char s[10];
int t,m,T;
int to[60];
int main()
{
    int i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&m,&T);
        to[0]=0;to[m+1]=0;to[m+2]=-1;
        for(i=1;i<=m;i++)
        {
            if(scanf("%d",&to[i])==0)
            {
                scanf("%s",s);
                to[i]=inf;
            }
        }
        memset(dp,0,sizeof(dp));
        dp[0][0]=1.0;
        for(j=0;j<=T;j++)
            for(i=0;i<=m;i++)
            {
               // printf("i %d j %d dp %lf\n",i,j,dp[i][j]);
                if(to[i+1]==inf)
                    dp[i+1][j+2]+=dp[i][j]*0.5;
                else
                    dp[i+1+to[i+1]][j+1]+=dp[i][j]*0.5;
                if(to[i+2]==inf)
                    dp[i+2][j+2]+=dp[i][j]*0.5;
                else
                    dp[i+2+to[i+2]][j+1]+=dp[i][j]*0.5;
            }
        for(i=0;i<T;i++)
            dp[m+1][T]+=dp[m+1][i];
        if(dp[m+1][T]==0.5)
            printf("Push. 0.5000\n");
        else if(dp[m+1][T]<0.5)
            printf("Bet against. %.4lf\n",dp[m+1][T]);
        else
            printf("Bet for. %.4lf\n",dp[m+1][T]);
    }
}