POJ2965 The Pilots Brothers' refrigerator 枚举 状态压缩

做法与1753相似。

代码如下：

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <algorithm>
#define START 0
#define END 65535
using namespace std;

char G[6][6];

int status, cpy, how[20], path[20], times;

void pre()
{
    how[1] = 4383, how[2] =8751, how[3] = 17487, how[4] = 34959, how[5] = 4593;
    how[6] = 8946, how[7] = 17652, how[8] = 35064, how[9] = 7953, how[10] = 12066;
    how[11] = 20292, how[12] = 36744, how[13] = 61713, how[14] = 61986;
    how[15] = 62532, how[16] = 63624;
    // 对每一个点进行反转所影响的区域的位压缩存储
}

bool dfs(int cur, int pos, int leavings)
{
    if (leavings == 0) {
    // 当组合排列完毕
        cpy = status;
        for (int i = 0; i < pos; ++i) {
            cpy ^= how[path[i]];
        }
        if (cpy == START) {
            printf("%d\n", times);
            for (int i = 0; i < pos; ++i) {
                printf("%d %d\n", (path[i]-1)/4+1, (path[i]-1)%4+1);
            }
            return true;
        }
        else {
            return false;
        }
    }
    else {
        for (int i = cur; i <= 16; ++i) {
            path[pos] = i;
            if (16-pos < leavings) {
                continue;
            }
            if (dfs(i+1, pos+1, leavings-1)) {
                return true;
            }
        }
        return false;
    }
}

bool OK(int times)
{
    if (dfs(1, 0, times)) {
        return true;
    }
    else {
        return false;
    }
}

int main()
{
    pre();
    // 读入数据
    for (int i = 1; i <= 4; ++i) {
        scanf("%s", G[i]+1);
        for (int j = 1; j <= 4; ++j) {
            G[i][j] = G[i][j] == '-' ? 0 : 1;
        }
    }
    
    for (int i = 4; i >= 1; --i) {
        for (int j = 4; j >= 1; --j) {
            status <<= 1;
            if (G[i][j]) {
                status |= 1;
                // 将整个图压缩到一个32位的数字中
            }
        }
    }
    
    bool finish = false;
    while (!finish) {
        if (OK(times)) {
            finish = true;
        }
        else {
            ++times;
        }
    }
    return 0;
}