POJ-2965 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;

}