poj 2411 Mondriaan's Dream(状态压缩dP)

题目：http://poj.org/problem?id=2411

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.

题意：一个矩阵，只能放1*2的木块，问将这个矩阵完全覆盖的不同放法有多少种。

思路:一道很经典的状态dp,但是还是很难想，横着放定义为11，竖着放定义为01.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
__int64 d[12][1<<12], f[1<<12];//用———int64,因为有可能数很大
int h, w, full;

bool ok(int x) //判断一行里是否出现连续奇数个1，第一行的话不允许
{
    int sum = 0;
    while(x>0)
    {
        if((x&1)==1)
        sum++;
        else
        {
            if((sum&1)==1)
            return false;
            sum = 0;
        }
        x = (x>>1);
    }
    if((sum&1)==1)
            return false;
    return true;
}
bool check(int x1, int x2)
{
    int xx = (1<<w)-1;
    if((x1|x2)==xx&&f[x1&x2])//去掉竖着00和竖着单独11的情况，一定不要忘了单独11的情况
    return true;
    return false;
}
int main()
{
    int i, j, k;
    memset(f, 0, sizeof(f));
    full = (1<<11);

    for(i = 0; i < full; i++)
    if(ok(i))
    f[i] = 1;   //记录连续的奇数个1
    while(~scanf("%d%d", &h, &w))
    {
        if(h==0&&w==0)
        break;
        full = (1<<w);
        memset(d, 0, sizeof(d));
        for(i = 0; i < full; i++)
        if(f[i])
        d[0][i] = 1;

        for(k = 1; k < h; k++)
        for(i = 0; i < full; i++)
        for(j = 0; j < full; j++)
        {
            if(check(i, j))
            d[k][i] += d[k-1][j];
        }
        printf("%I64d
", d[h-1][full-1]); //最后一行，而且满1
    }
    return 0;
}