标签：following tab NPU too clu 超过 == color git
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 8903 | Accepted: 3772 |
Description
Input
Output
Sample Input
2 2 2
0.9 0.9
1 0.9
0 0 0
Sample Output
0.972
Source
Solution
概率DP
定义$dp[i][j][k]$表示第$i$个队伍做了$j$道题对了$k$道的概率。转移方程显然。
再定义$s[i][j]$表示第$i$个队对了不超过$j$道题的概率，$s[i][j]=\sum{dp[i][t][k]},0<=k<=j$
然后所有队伍都对了至少一道题的概率就是$p1=\prod{(1-s[i][0])}$
因为冠军队至少对了$n$道，那么最后的概率应该是$p1$减去每个队伍都没有达到$n$的概率：$\prod{s[i][n-1]}$，在这里要注意，还要保证每个队伍都至少对了1道题，因为是递推转移得到$s$，不能把$0$的贡献算进去，所以$s$从2开始递推。
Code
#include<iostream> #include<cstdio> using namespace std; double dp[1005][35][35], s[1005][35], p[1005][35]; int m, t, n; int main() { while(scanf("%d%d%d", &m, &t, &n) != EOF) { if(m == 0 && t == 0 && n == 0) break; for(int i = 1; i <= t; i ++) for(int j = 1; j <= m; j ++) scanf("%lf", &p[i][j]); for(int i = 1; i <= t; i ++) { dp[i][0][0] = 1; for(int j = 1; j <= m; j ++) { dp[i][j][0] = dp[i][j - 1][0] * (1 - p[i][j]); for(int 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]); } for(int j = 0; j <= m; j ++) s[i][j] = dp[i][m][j]; } for(int i = 1; i <= t; i ++) for(int j = 2; j <= m; j ++) s[i][j] = s[i][j] + s[i][j - 1]; double p1 = 1; for(int i = 1; i <= t; i ++) p1 *= (1 - s[i][0]); double p2 = 1; for(int i = 1; i <= t; i ++) p2 *= s[i][n - 1]; if(n == 1) p2 = 0; printf("%0.3lf\n", p1 - p2); } return 0; }
【POJ】2151：Check the difficulty of problems【概率DP】
标签：following tab NPU too clu 超过 == color git
原文地址：https://www.cnblogs.com/wans-caesar-02111007/p/9835722.html
