zoj3822||牡丹江现场赛D题 概率dp
来源:互联网 发布:知乎v领毛衣配衬衫女 编辑:程序博客网 时间:2024/05/24 05:02
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5376
Edward is the headmaster of Marjar University. He is enthusiastic about chess and often plays chess with his friends. What's more, he bought a large decorative chessboard with N rows and M columns.
Every day after work, Edward will place a chess piece on a random empty cell. A few days later, he found the chessboard was dominated by the chess pieces. That means there is at least one chess piece in every row. Also, there is at least one chess piece in every column.
"That's interesting!" Edward said. He wants to know the expectation number of days to make an empty chessboard of N × M dominated. Please write a program to help him.
Input
There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:
There are only two integers N and M (1 <= N, M <= 50).
Output
For each test case, output the expectation number of days.
Any solution with a relative or absolute error of at most 10-8 will be accepted.
Sample Input
21 32 2
Sample Output
3.0000000000002.666666666667
/**zoj 3822题意:问把一个n*m的棋盘填成每行至少一个棋子,每列至少一个棋子的状态,需要棋子的数学期望解题思路: dp[i][j][k]表示到达结果状态时所需要的棋子数的期望值。那么dp[0][0][0]就是结果,状态转移方程见代码。*/#include <stdio.h>#include <iostream>#include <string.h>#include <algorithm>using namespace std;int T,n,m;double dp[55][55][2555];int main(){ scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); memset(dp,0,sizeof(dp)); for(int i=n; i>=0; i--) { for(int j=m; j>=0; j--) { if(j==m&&n==i) continue; for(int t=i*j; t>=max(i,j); t--) { dp[i][j][t]+=dp[i][j][t+1]*1.0*(i*j-t)/(n*m-t); dp[i][j][t]+=dp[i][j+1][t+1]*1.0*(m-j)*i/(n*m-t); dp[i][j][t]+=dp[i+1][j][t+1]*(1.0)*(n-i)*j/(n*m-t); dp[i][j][t]+=dp[i+1][j+1][t+1]*(1.0)*(n-i)*(m-j)/(n*m-t); dp[i][j][t]+=1.0; } } } printf("%.12lf\n",dp[0][0][0]); } return 0;}
- zoj3822 Domination 牡丹江现场赛D题 概率DP
- zoj3822||牡丹江现场赛D题 概率dp
- ZOJ3822 ACM-ICPC 2014 亚洲区域赛牡丹江赛区现场赛D题Domination 概率DP
- ZOJ3822 ACM-ICPC 2014 亚洲区域赛牡丹江赛区现场赛D题Domination 概率DP(两种解法)
- 2014牡丹江区域赛D(概率DP)ZOJ3822
- ZOJ 3822 Domination 概率DP 2014年ACM_ICPC亚洲区域赛牡丹江现场赛D题
- 2014牡丹江现场赛A题D题I题(水,概率Dp,水)ZOJ 3819,3822,3827
- ZOJ 3822 Domination(概率dp 牡丹江现场赛)
- ZOJ 3822Domination /2014牡丹江现场赛D题 (dp)
- zoj 3822 Domination 概率dp 2014牡丹江站D题
- zoj3822(概率DP)
- ZOJ3822-Domination 概率DP
- ZOJ3822(概率DP)
- zoj3822(概率dp)
- zoj3822(概率dp)
- zoj3822-概率dp-Domination
- zoj3822 Domination (The 2014 ACM-ICPC Asia Mudanjiang Regional Contest D题)概率dp
- ZOJ 3826 Hierarchical Notation(2014 牡丹江 D,概率DP)
- Android中根据联系人的Recipient ID获取电话号码
- app插件化-绑定生命周期版本
- 二进制流和文本流区别(C语音)
- mongoDB的自动递增
- C++模板编程实现二维数组
- zoj3822||牡丹江现场赛D题 概率dp
- mysql基本操作
- intent传递
- vdsm的SSL证书验证过程
- iOS修改相机cancel键为中文方法,系统相机,系统相册
- android控件总结
- POJ 3071 Football 概率DP
- JVM统计监控工具-jstat
- python实现获取文件列表中每个文件关键字