ZOJ 3822(求期望)
Domination
Time Limit: 8 Seconds Memory Limit: 131072 KB Special Judge
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
2
1 3
2 2
Sample Output
3.000000000000
2.666666666667
Author: JIANG, Kai
Source: The 2014 ACM-ICPC Asia Mudanjiang Regional Contest
做了这么多概率dp,结果这道还是没做出来,心情已经不能用郁闷二字来形容了。。。
一、直接求期望
首先是状态的问题,一直在用二维,其实在算概率的时候就应该意识到二维的概率好难算,数据又是50的,很明显要用三维啊!真是笨死了!
dp[i][j][k]代表走了k步,已经有i行,j列安放了棋子。
接下来就是算概率的问题
很明显有四种可转移状态:
1、dp[i][j][k+1]表示走完k+1步,仍是有i行,j列安放了棋子。即安放的第k+1个棋子在i,j所占据的区域,概率为 (i * j - k) / (n * m - k);
2、dp[i][j+1][k+1]表示走完k+1步,有i行,j + 1列安放了棋子。即安放的第k+1个棋子在i行中但不在j列,概率为 i * (m - j) / (n * m - k);
3、dp[i+1][j][k+1]表示走完k+1步,有i + 1行,j列安放了棋子。即安放的第k+1个棋子在j列中但不在i行,概率为 (n - i) * j / (n * m - k);
4、dp[i+1][j+1][k+1]表示走完k+1步,有i + 1行,j + 1列安放了棋子。即安放的第k+1个棋子既不在j列中也不在i行,概率为 (n - i) * (m - j) / (n * m - k);
然后是初始化问题
当i == n && j == m时候,dp[i][j][k] = 0;
无意义的状态全部初始为零,没有影响。
#include <cstdio>
#include <iostream>
#include <sstream>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;
#define ll long long
#define _cle(m, a) memset(m, a, sizeof(m))
#define repu(i, a, b) for(int i = a; i < b; i++)
#define repd(i, a, b) for(int i = b; i >= a; i--)
#define sfi(n) scanf("%d", &n)
#define sfl(n) scanf("%lld", &n)
#define pfi(n) printf("%d\n", n)
#define pfl(n) printf("%lld\n", n)
#define MAXN 55
double dp[MAXN][MAXN][MAXN * MAXN]; int main()
{
int T;
sfi(T);
while(T--)
{
int n, m;
sfi(n), sfi(m);
_cle(dp, );
for(int i = n; i >= ; i--)
for(int j = m; j >= ; j--)
for(int k = i * j; k >= max(i, j); k--)
{
if(n == i && j == m) continue;
dp[i][j][k] += (dp[i][j][k + ] + 1.0) * 1.0 * (i * j - k);
dp[i][j][k] += (dp[i][j + ][k + ] + 1.0) * 1.0 * (i * (m - j));
dp[i][j][k] += (dp[i + ][j][k + ] + 1.0) * 1.0 * (j * (n - i));
dp[i][j][k] += (dp[i + ][j + ][k + ] + 1.0) * 1.0 * ((n - i) * (m - j));
dp[i][j][k] = dp[i][j][k] / (1.0 * (n * m - k));
//printf("%d %d %d : %.12lf\n", i, j, k, dp[i][j][k]);
}
printf("%.12lf\n", dp[][][]);
}
return ;
}
二、先求概率,再求期望
dp[i][j][k]表示放k个棋子达到有i行,j列安放了棋子的概率。
这里有一大误区,就是开始我和我队友都搞错了,结果还以为自己是算的不对,其实是思考错了,就是在算期望是应只考虑真正起作用的棋子,eg:
2 2
dp[2][2][3] = 1, dp[2][2][4] = 1;
其实在放第四颗棋子时就已经必然为两行两列,第四颗棋子已不起作用,放与不放无关痛痒,所以,放第k颗棋子的概率为:
dp[i][j][k] - dp[i][j][k - 1];
#include <cstdio>
#include <iostream>
#include <sstream>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;
#define ll long long
#define _cle(m, a) memset(m, a, sizeof(m))
#define repu(i, a, b) for(int i = a; i < b; i++)
#define repd(i, a, b) for(int i = b; i >= a; i--)
#define sfi(n) scanf("%d", &n)
#define sfl(n) scanf("%lld", &n)
#define pfi(n) printf("%d\n", n)
#define pfl(n) printf("%lld\n", n)
#define MAXN 55
double dp[MAXN][MAXN][MAXN * MAXN]; int main()
{
int T;
sfi(T);
while(T--)
{
int n, m;
sfi(n), sfi(m);
_cle(dp, );
dp[][][] = 1.0;
repu(i, , n + )
repu(j, , m + )
repu(k, max(i, j), i * j + )
{
if(k < ) continue;
dp[i][j][k] += dp[i][j][k - ] * 1.0 * (i * j - k + );
dp[i][j][k] += dp[i - ][j][k - ] * 1.0 * ((n - i + ) * j);
dp[i][j][k] += dp[i][j - ][k - ] * 1.0 * (i * (m - j + ));
dp[i][j][k] += dp[i - ][j - ][k - ] * 1.0 * ((n - i + ) * (m - j + ));
dp[i][j][k] /= (1.0 * (n * m - k + ));
//printf("%d %d %d : %.12lf\n", i, j, k, dp[i][j][k]);
}
double ans = 0.0;
repu(i, max(n, m), n * m + )
ans += (dp[n][m][i] - dp[n][m][i - ]) * 1.0 * i;
printf("%.12lf\n", ans);
//double f = ans * 3.0; }
return ;
}
ZOJ 3822(求期望)的更多相关文章
- ZOJ 3822 Domination 期望dp
Domination Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.zju.edu.cn/onlinejudge/showProblem ...
- zoj 3822 Domination(dp)
题目链接:zoj 3822 Domination 题目大意:给定一个N∗M的棋盘,每次任选一个位置放置一枚棋子,直到每行每列上都至少有一枚棋子,问放置棋子个数的期望. 解题思路:大白书上概率那一张有一 ...
- HDU4870_Rating_双号从零单排_高斯消元求期望
原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=4870 原题: Rating Time Limit: 10000/5000 MS (Java/Other ...
- sgu 495. Kids and Prizes (简单概率dp 正推求期望)
题目链接 495. Kids and Prizes Time limit per test: 0.25 second(s)Memory limit: 262144 kilobytes input: s ...
- ZOJ 3822 Domination
题意: 一个棋盘假设每行每列都有棋子那么这个棋盘达到目标状态 如今随机放棋子 问达到目标状态的期望步数 思路: 用概率来做 计算第k步达到目标状态的概率 进而求期望 概率计算方法就是dp ...
- HDU3853-LOOPS(概率DP求期望)
LOOPS Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others) Total Su ...
- HDU 5159 Card (概率求期望)
B - Card Time Limit:5000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Sta ...
- Poj 2096 (dp求期望 入门)
/ dp求期望的题. 题意:一个软件有s个子系统,会产生n种bug. 某人一天发现一个bug,这个bug属于某种bug,发生在某个子系统中. 求找到所有的n种bug,且每个子系统都找到bug,这样所要 ...
- poj 2096 Collecting Bugs 【概率DP】【逆向递推求期望】
Collecting Bugs Time Limit: 10000MS Memory Limit: 64000K Total Submissions: 3523 Accepted: 1740 ...
随机推荐
- JavaScript经典代码【二】【javascript判断用户点了鼠标左键还是右键】
IE 下 onMouseDown 事件有个 events.button 可以返回一个数值,根据数值判断取得用户按了那个鼠标键 events.button==0 默认.没有按任何按钮. events.b ...
- CALayer总结(三)
CPU VS GPU 动画和屏幕上组合的图层实际上被一个单独的进程管理,而不是你的应用程序.这个进程就是所谓的渲染服务.在iOS5和之前的版本是SpringBoard进程(同时管理着iOS的主屏).在 ...
- Python-爬虫初学
#爬取网站中的图片 1 import re #正则表达式库 import urllib #url链接库 def getHtml(url): page = urllib.urlopen(url) #打开 ...
- Struts BaseAction工具类,封装Session,Request,Application,ModelDriven
package com.ssh.shop.action; import java.io.InputStream; import java.lang.reflect.ParameterizedType; ...
- 小型网站如何防范DDoS攻击
ddos(Distributed Denial of Service,分布式拒绝服务攻击),俗称洪水攻击.是在传统的DoS攻击基础之上产生的新的破坏力更强的攻击方式.分布式拒绝服务攻击是指借助于客户/ ...
- mysql delimiter
默认情况下,mysql遇到分号; 就认为是一个命令的终止符, 就会执行命令.而有些时候,我们不希望这样,比如存储过程中包含多个语句,这些语句以分号分割,我们希望这些语句作为一个命令,一起执行,怎么解决 ...
- Echarts个人实例
1.deviceOperateTrendIndex.jsp <%@ page language="java" contentType="text/html; cha ...
- JavaWeb学习总结(七)—HttpServletRequest
一.Request概述 request是Servlet.service()方法的一个参数,类型为javax.servlet.http.HttpServletRequest.在客户端发出每个请求时,服务 ...
- Nginx反向代理负载均衡
环境准备: 总共四台机器,两台装有Nginx的机器做负载均衡,两台机器装有Apache作为WEB服务器. 机器信息 hostname IP 说明 lb01 192.168.1.19 nginx主负载均 ...
- wireshark使用教程
Wireshark: https://www.wireshark.org/ 安装: apt-get install wireshark 教程: http://blog.csdn.net/leichel ...