POJ 2411 Mondriaan's Dream（插头DP，轮廓线）

最新推荐文章于 2020-03-14 20:52:14 发布

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最新推荐文章于 2020-03-14 20:52:14 发布

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本文介绍了一种使用轮廓线动态规划(DP)的方法来解决如何用1×2的小矩形填充n×m的大矩形的问题。通过状态转移实现不同大小矩形的填充方案计数，特别适用于计算机科学领域的算法设计与分析。

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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

Sample Output

题意：用1*2的小方格填满n*m的方案数

分析：轮廓线DP，由一格一格状态转化

代码；

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <queue>
#define mem(p,k) memset(p,k,sizeof(p));
#define rep(a,b,c) for(int a=b;a<c;a++)
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
const int maxn=10;
int n,m,ans;
ll dp[2][100000];
int main()
{
    while(~scanf("%d%d",&n,&m) && n+m){
        if(n*m&1){
            cout<<0<<endl;continue;
        }
        if(n<m)swap(n,m);
        mem(dp,0);
        int now=0;
        dp[now][(1<<m)-1]=1;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                now^=1;
                mem(dp[now],0);
                for(int k=0;k<(1<<m);k++){
                    if(dp[now^1][k]){
                        dp[now][k^(1<<j)]+=dp[now^1][k];
                        if(j && k&(1<<j) && !(k&1<<j-1)){
                            dp[now][k^1<<j-1]+=dp[now^1][k];
                        }
                    }
                }
            }
        }
        cout<<dp[now][(1<<m)-1]<<endl;
    }

    return 0;
}