POJ 2411 Mondriaan's Dream

题目链接：http://poj.org/problem?id=2411

状态压缩Dynamic Programming. For each row, at ith position, 1 means that there is a block placed at this row and next row (vertically). otherwise, its 0. For the example in question, the state of

For the example in question, the state of first row is (00100001100) binary. The state of second row is (11010010000). and so on. The state of last row must be all 0 (has no influence in next row).

We use dp[i][j] to record the number of possible ways to place row i with state j. The final answer is dp[h][0]. In the initialization, all values in dp is 0 and dp[0][0]=1. if row i with state A matches next row j with state B, dp[j][B]+=dp[i][A].

代码如下：

#include <iostream>
#include <math.h>
#include <stdio.h>
#include <cstdio>
#include <algorithm>
#include <string.h>
#include <string>
#include <sstream>
#include <cstring>
#include <queue>
#include <vector>
#include <functional>
#include <cmath>
#include <set>
#define SCF(a) scanf("%d", &a)
#define IN(a) cin>>a
#define FOR(i, a, b) for(int i=a;i<b;i++)
#define Infinity 999999999
typedef long long Int;
using namespace std;

int h, w; //1 <= h, w <= 11
          //max states are 2^11 = 2048
bool match[2048][2048];
Int state[12][2048]; //i -> row, j -> state
int states; //num of states for a row

bool check(int a, int b)
{
    if (a&b)
        return false;
    int c = a | b;
    int zeroCon = 0;
    FOR(i, 0, w)
    {
        if (!(c & (1 << i)))
            zeroCon++;
        else
        {
            if (zeroCon % 2)
                return false;
            else
                zeroCon = 0;
        }
    }
    if (zeroCon % 2)
        return false;
    return true;
}

int main()
{
    FOR(i, 0, 12)
    {
        FOR(j, 0, 2048)
            state[i][j] = -1;
    }

    while (scanf("%d %d", &h, &w) != EOF)
    {
        if (h == 0 && w == 0)
            break;
        if ((h*w) % 2)
        {
            printf("0\n");
            continue;
        }

        if (h < w)
        {
            int temp = w;
            w = h;
            h = temp;
        }
        
        states = (1 << w); // 0 - states-1
        
        FOR(i, 0, states)
        {
            FOR(j, 0, states)
            {
                match[i][j] = check(i, j);
            }
        }

        FOR(i, 0, h + 1)
        {
            FOR(j, 0, states)
                state[i][j] = 0;
        }

        state[0][0] = 1;

        FOR(i, 1, h + 1)
        {
            FOR(si, 0, states)
            {
                FOR(sp, 0, states)
                {
                    if(match[sp][si]==true)
                        state[i][si] += state[i - 1][sp];
                }
            }
        }

        printf("%lld\n", state[h][0]);
    }
    return 0;
}