POJ 2151 Check the difficulty of problems 概率DP

题目大意：

就是现在有一场比赛有M道题(<= 30), 一共有T个队伍参赛( <= 1000), 现在知道每只队伍做出每道题的概率，问比赛结束之后所有队伍都有过题且过题最多的队伍过了N题或以上的概率( N <= M)

大致思路：

就是一个典型的dp吧 ( ﹁ ﹁ ), 过程见代码注释

Result  :  Accepted     Memory  :  15996 KB     Time  :  125 ms

/* * Author: Gatevin * Created Time:  2014/12/3 14:44:18 * File Name: Asuna.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;/* * 用dp[i][j][k]表示第i个队伍在前j道题中做出k道德概率，那么 * dp[i][j][k] = dp[i][j - 1][k - 1]*p[i][j] + dp[i][j - 1][k]*(1 - p[i][j]) 0 < k < j * dp[i][j][k] = dp[i][j - 1][k - 1]*p[i][j] k == j * dp[i][j][k] = dp[i][j - 1][k]*(1 - p[i][j]) k == 0 * 那么这样可以知道每个队伍做出的题目在0~m之间的概率 * 用sp[i]表示第i个队伍做出1 ~ n - 1道题的概率 sp[i] = ∑dp[i][m][1 ~ n - 1] * 用s[i][j]表示前i个队伍中有j个做出了n题或以上的概率 * 则s[i][j] = s[i - 1][j]*sp[i] + s[i - 1][j - 1]*(1 - sp[i] - dp[i][m][0]) 0 < j < i *   s[i][j] = s[i - 1][j]*sp[i] j == 0 *   s[i][j] = s[i - 1][j - 1]*(1 - sp[i] - dp[i][m][0]) j == i * 最后的结果就是∑s[t][1~t] *///这题没有卡精度，看见POJ2151的discuss里面好多人因为.lf 和.f 的问题发了好多帖子,表示想起某只队友...int m, t, n;double dp[1010][31][31];double p[1010][31];double s[1010][1010];double sp[1010];int main(){    while(scanf("%d %d %d", &m, &t, &n), n || t || m)//刚开始n和m打反了交了一发WA...囧    {        for(int i = 1; i <= t; i++)            for(int j = 1; j <= m; j++)                scanf("%lf", &p[i][j]);        memset(dp, 0, sizeof(dp));        for(int i = 1; i <= t; i++)        {            dp[i][0][0] = 1;            for(int j = 1; j <= m; j++)                for(int k = 0; k <= j; k++)                {                    if(k == 0) dp[i][j][k] = dp[i][j - 1][k]*(1 - p[i][j]);                    else if(k == j) dp[i][j][k] = dp[i][j - 1][k - 1]*p[i][j];                    else dp[i][j][k] = dp[i][j - 1][k - 1]*p[i][j] + dp[i][j - 1][k]*(1 - p[i][j]);                    //cout<<dp[i][j][k]<<" dp["<<i<<"]["<<j<<"]["<<k<<"] "<<endl;                }        }        memset(sp, 0, sizeof(sp));        for(int i = 1; i <= t; i++)            for(int k = 1; k < n; k++)                sp[i] += dp[i][m][k];        memset(s, 0, sizeof(s));        s[0][0] = 1;        for(int i = 1; i <= t; i++)            for(int j = 0; j <= i; j++)            {                if(j == 0) s[i][j] = s[i - 1][j]*sp[i];                else if(j == i) s[i][j] = s[i - 1][j - 1]*(1 - sp[i] - dp[i][m][0]);                else s[i][j] = s[i - 1][j]*sp[i] + s[i - 1][j - 1]*(1 - sp[i] - dp[i][m][0]);            }        double ans = 0;        for(int i = 1; i <= t; i++)            ans += s[t][i];        printf("%.3f\n", ans);    }    return 0;}