zoukankan      html  css  js  c++  java
  • POJ 2411 Mondriaan's Dream( 轮廓线dp )

    最普通的轮廓线dp... 复杂度O(nm2min(n, m))

    --------------------------------------------------------------------

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
     
    using namespace std;
     
    typedef long long ll;
     
    #define b(x) (1 << (x))
     
    const int maxn = 13;
     
    ll dp[2][b(maxn)];
     
    int main() {
    int n, m;
    while(scanf("%d%d", &n, &m) == 2 && n && m) {
    if(n < m) swap(n, m);
    int c = 0, p = 1;
    memset(dp, 0, sizeof dp);
    dp[c][b(m) - 1] = 1;
    for(int i = 0; i < n; i++) 
       for(int j = 0; j < m; j++) {
        swap(c, p);
        memset(dp[c], 0, sizeof dp[c]);
           for(int s = b(m); s--; ) {
            if(s & b(m - 1)) {
       dp[c][s << 1 ^ b(m)] += dp[p][s];
       if(j && !(s & 1)) dp[c][s << 1 ^ b(m) | 3] += dp[p][s];
    } else if(i) 
       dp[c][s << 1 | 1] += dp[p][s];
           }
       }
    printf("%lld ", dp[c][b(m) - 1]);
    }
    return 0;
    }

    -------------------------------------------------------------------- 

    Mondriaan's Dream
    Time Limit: 3000MSMemory Limit: 65536K
    Total Submissions: 13390Accepted: 7802

    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!

    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 2 1 3 1 4 2 2 2 3 2 4 2 11 4 11 0 0 

    Sample Output

    1 0 1 2 3 5 144 51205 

    Source

  • 相关阅读:
    07.消除过期对象的引用
    1.1进程和多线程概述
    1.2什么是操作系统
    06.避免创建不必要的对象
    05.依赖注入优先于硬连接资源
    04.使用私有构造器执行非实例化
    03.使用私有构造方法或枚类实现 Singleton 属性
    02.当构造参数过多时使用builder模式
    01.考虑使用静态工厂方法替代构造方法
    iiS申请地址
  • 原文地址:https://www.cnblogs.com/JSZX11556/p/4769720.html
Copyright © 2011-2022 走看看