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  • Check the difficulty of problems(概率+DP)

    http://poj.org/problem?id=2151

    看的题解。。表示没看懂状态转移方程。。

    #include<stdio.h>
    #include<string.h>
    int m,t,n;
    double dp[1002][32][32],p[1002][32],tt[1002][32];
    int main()
    {
        int i,j,k;
        while(~scanf("%d%d%d",&m,&t,&n))
        {
            if (!m && !t && !n)
                break;
            memset(dp,0.0,sizeof(dp));
            memset(tt,0.0,sizeof(tt));
            for (i = 1; i <= t; i ++)
                for (j = 1; j <= m; j ++)
                {
                    scanf("%lf",&p[i][j]);
                }
    
            for (i = 1; i <= t; i ++)
            {
                dp[i][0][0] = 1.0;
                for (j = 1; j <= m; j ++)
                {
                    dp[i][j][0] = dp[i][j-1][0]*(1-p[i][j]);
                }
                for (j = 1; j <= m; j ++)
                {
                    for (k = 1; k <= j; k ++)
                    {
                        dp[i][j][k] = dp[i][j-1][k-1]*p[i][j] + dp[i][j-1][k]*(1-p[i][j]);
                    }
                }
                tt[i][0] = dp[i][m][0];
                for (k = 1; k <= m; k ++)
                {
                    tt[i][k] = tt[i][k-1] + dp[i][m][k];
                }
            }
            double p1 = 1.0,p2=1.0;
            for (i = 1; i <= t; i ++)
            {
                p1 *= (tt[i][m]-tt[i][0]);
                p2 *= (tt[i][n-1]-tt[i][0]);
            }
            printf("%.3f
    ",p1-p2);
        }
        return 0 ;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/lahblogs/p/3268903.html
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