POJ2411:Mondriaan's Dream(状态压缩)
来源:互联网 发布:董先生知乎 编辑:程序博客网 时间:2024/05/21 11:06
Description
Squares and rectangles fascinated the famous Dutch painter Piet Mondriaan. One night, after producing the drawings in his 'toilet series' (where he had to use his toilet paper to draw on, for all of his paper was filled with squares and rectangles), he dreamt of filling a large rectangle with small rectangles of width 2 and height 1 in varying ways.
Expert as he was in this material, he saw at a glance that he'll need a computer to calculate the number of ways to fill the large rectangle whose dimensions were integer values, as well. Help him, so that his dream won't turn into a nightmare!
Expert as he was in this material, he saw at a glance that he'll need a computer to calculate the number of ways to fill the large rectangle whose dimensions were integer values, as well. Help him, so that his dream won't turn into a nightmare!
Input
The input contains several test cases. Each test case is made up of two integer numbers: the height h and the width w of the large rectangle. Input is terminated by h=w=0. Otherwise, 1<=h,w<=11.
Output
For each test case, output the number of different ways the given rectangle can be filled with small rectangles of size 2 times 1. Assume the given large rectangle is oriented, i.e. count symmetrical tilings multiple times.
Sample Input
1 21 31 42 22 32 42 114 110 0
Sample Output
10123514451205
用2进制的01表示不放还是放第i行只和i-1行有关枚举i-1行的每个状态,推出由此状态能达到的i行状态如果i-1行的出发状态某处未放,必然要在i行放一个竖的方块,所以我对上一行状态按位取反之后的状态就是放置了竖方块的状态。然后用搜索扫一道在i行放横着的方块的所有可能,并且把这些状态累加上i-1的出发状态的方法数,如果该方法数为0,直接continue。
#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int n,m;__int64 dp[15][1<<11],tem;void dfs(int i,int p,int k){ if(k>=m) { dp[i][p]+=tem; return; } dfs(i,p,k+1); if(k<=m-2 && !(p&1<<k) && !(p&1<<k+1)) dfs(i,p|1<<k|1<<k+1,k+2);}int main(){ int i,j; while(~scanf("%d%d",&n,&m),n+m) { memset(dp,0,sizeof(dp)); tem = 1; dfs(1,0,0); for(i = 2; i<=n; i++) { for(j = 0; j<1<<m; j++) { if(dp[i-1][j]) tem = dp[i-1][j]; else continue; dfs(i,~j&((1<<m)-1),0); } } printf("%I64d\n",dp[n][(1<<m)-1]); } return 0;}
4 0
- poj2411 Mondriaan's Dream 状态压缩dp
- POJ2411:Mondriaan's Dream(状态压缩)
- poj2411 Mondriaan's Dream 状态压缩dp
- 【poj2411】Mondriaan's Dream 状态压缩dp
- poj2411——Mondriaan's Dream(状态压缩DP)
- POJ2411 Mondriaan's Dream(状态压缩DP)
- 【POJ2411】Mondriaan's Dream-状态压缩DP(插头DP?)
- POJ2411 Mondriaan's Dream(压缩DP)
- POJ2411-Mondriaan's Dream-状态dp
- POJ2411 Mondriaan's Dream
- POJ2411--Mondriaan's Dream
- POJ2411 Mondriaan's Dream
- poj2411 Mondriaan's Dream
- Mondriaan's Dream POJ2411
- poj2411: Mondriaan's Dream
- 【POJ2411】Mondriaan's Dream
- Mondriaan's Dream(DP-之状态压缩的动态规划poj2411)
- zoj 1100&& poj2411 Mondriaan's Dream (dfs&&枚举&&状态压缩dp)好题
- Idea Tomcat sysout输出中文乱码问题解法
- JavaScript中的原型和继承
- java模拟真实用户上传文件
- mysql 与java时间类型
- OWASP WebGoat---安全测试学习笔记(六)---代码质量
- POJ2411:Mondriaan's Dream(状态压缩)
- 9-5-一元一次方程类
- 安装redis总结
- Mariadb源码编译过程
- oracle笔记2-sql查询
- 大学毕业后拉开差距的真正原因
- 变位词的实现与理解
- 企业建站系统MiinCMP1.0.5 版发布!
- HttpServletRequest.getParameter(String)和getAttribute()的区别