POJ 2965 The Pilots Brothers' refrigerator 状态压缩+广搜

题意
4*4个开关，每当点击某个开关，该开关及其所在行和所在列的所有开关都会发生改变。给出一组开关状态，求把这组状态变为全开所需的最少步数及相关操作。

样例输入
-+--
----
----
-+--

样例输出
6
1 1
1 3
1 4
4 1
4 3
4 4

思路
状态压缩，BFS暴搜

#include <cstdio>
#include <string.h>
#include <queue>
using namespace std;
const int flip[16] = {0xF888, 0xF444, 0xF222, 0xF111, 0x8F88, 0x4F44, 0x2F22, 0x1F11, 0x88F8, 0x44F4, 0x22F2, 0x11F1, 0x888F, 0x444F, 0x222F, 0x111F};
const int N = 65536;
int cnt[N], path[N], change[N];

void Print(int x)
{
    if(x)
    {
        Print(path[x]);
        printf("%d %d
", change[x]/4+1, change[x]%4+1);
    }
}

void BFS(int x)
{
    bool vis[N] = {0};
    memset(cnt, -1, sizeof(cnt));
    vis[0] = true;
    cnt[0] = 0;
    queue<int> q;
    q.push(0);
    int top, temp;
    while(!q.empty())
    {
        top = q.front();
        for(int i=0; i<16; ++i)
        {
            temp = top ^ flip[i];
            if(!vis[temp])
            {
                cnt[temp] = cnt[top] + 1;
                path[temp] = top;
                change[temp] = i;
                if(temp == x)
                {
                    printf("%d
", cnt[x]);
                    Print(x);
                    return ;
                }
                vis[temp] = true;
                q.push(temp);
            }
        }
        q.pop();
    }
}

int main(void)
{
    int x = 0;
    for(int i=0; i<4; ++i)
    {
        for(int j=0; j<4; ++j)
            x = (x << 1) + (getchar() == '+' ? 1 : 0);
        getchar();
    }
    BFS(x);
    return 0;
}

后来看了这个题解，恍然大悟，又优化了一下代码

http://www.cnblogs.com/Java-tp/p/3873557.html

#include <cstdio>
const int flip[] = {0xF888, 0xF444, 0xF222, 0xF111, 0x8F88, 0x4F44, 0x2F22, 0x1F11, 0x88F8, 0x44F4, 0x22F2, 0x11F1, 0x888F, 0x444F, 0x222F, 0x111F};
int main(void)
{
    int x = 0, num = 0;
    for(int i=0; i<16; --i)
        for(int j=0; j<5; x ^= getchar() == '+' ? flip[i] : 0, ++j, ++i);
    for(int temp = x; temp; num += temp&1, temp>>=1);
    printf("%d
", num);
    for(int i=15; ~i; --i, x>>=1)
        if(x&1)
            printf("%d %d
", (i>>2)+1, (i&3)+1);
}