poj 3279 Fliptile （状压）

题目链接：http://poj.org/problem?id=3279

Fliptile Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7831   Accepted: 2935 Description Farmer John knows that an intellectually satisfied cow is a happy cow who will give more milk. He has arranged a brainy activity for cows in which they manipulate an M × N grid (1 ≤ M ≤ 15; 1 ≤ N ≤ 15) of square tiles, each of which is colored black on one side and white on the other side. As one would guess, when a single white tile is flipped, it changes to black; when a single black tile is flipped, it changes to white. The cows are rewarded when they flip the tiles so that each tile has the white side face up. However, the cows have rather large hooves and when they try to flip a certain tile, they also flip all the adjacent tiles (tiles that share a full edge with the flipped tile). Since the flips are tiring, the cows want to minimize the number of flips they have to make. Help the cows determine the minimum number of flips required, and the locations to flip to achieve that minimum. If there are multiple ways to achieve the task with the minimum amount of flips, return the one with the least lexicographical ordering in the output when considered as a string. If the task is impossible, print one line with the word "IMPOSSIBLE". Input Line 1: Two space-separated integers:  M and  N  Lines 2.. M+1: Line  i+1 describes the colors (left to right) of row i of the grid with  N space-separated integers which are 1 for black and 0 for white Output Lines 1.. M: Each line contains  N space-separated integers, each specifying how many times to flip that particular location. Sample Input 4 4 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 Sample Output 0 0 0 0 1 0 0 1 1 0 0 1 0 0 0 0 Source USACO 2007 Open Silver

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题目大意：给出一个地图，上面有 黑（1）白（0）砖，反转一块砖， 可以连带着将它的上下左右都反转，黑反转成白， 白变成黑，

                 问你用最少的反转次数，使得所有砖都变成  白砖，并输出，如果有多种方案，输出字典序最小的

解析：枚举第一行所有情况，第一行确定后，后面都定了， 从00000.....到11111....如果上一行当前位置为黑，则它的正下方必须反转，

          最后判断最后一行是否全为 0，  是 0 就是这种方案可行

          借鉴大牛的博客：http://blog.youkuaiyun.com/HelloWorld10086/article/details/45874945

代码如下：

#include<iostream>
#include<algorithm>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<string>
#include<cstdio>
#include<cstring>
#include<cctype>
#include<cmath>
#define N 19
using namespace std;
const int inf = 1e9;
const int mod = 1<<30;
const double eps = 1e-8;
const double pi = acos(-1.0);
typedef long long LL;
int mp[N][N], g[N][N], tm[N][N], rec[N][N];
int m, n, ans;
int nx[6] = {-1, 0, 0, 1, 0};
int ny[6] = {0, -1, 1, 0, 0};

int check()  // 判断是否满足条件
{
    for(int i = 1; i <= n; i++)
        if(g[m][i] == 1) return 0;
    return 1;
}

void flip(int x, int y)  // 改变状态，反转x, y 后 g[N][N] 数组的状态
{
    int u, v;
    tm[x][y] = 1;
    for(int i = 0; i <= 4; i++)
    {
         u = x + nx[i];
         v = y + ny[i];
         g[u][v] = !g[u][v];
    }
}

void solve(int s)
{
    int i, j, cnt = 0;
    memset(tm, 0, sizeof(tm)); //标记是否反转过
    memcpy(g, mp, sizeof(mp)); //重复操作
    for(i = 0; i < n; i++)
    {
        if((s >> i) & 1)
            flip(1, i + 1), cnt++;
    }

    for(i = 2; i <= m; i++)
    for(j = 1; j <= n; j++)
        if(g[i - 1][j])  flip(i, j), cnt++;

    if(check() && cnt < ans)
    {
        memcpy(rec, tm, sizeof(tm));
        ans = cnt;
    }
}


int main()
{
    int i, j;
    scanf("%d%d", &m, &n);

    for(i = 1; i <= m; i++)
    for(j = 1; j <= n; j++)
       scanf("%d", &mp[i][j]);
    int k = 1 << n;
    ans = inf;
    for(i = 0; i <= k; i++) // 枚举所以状态
        solve(i);

    if(ans != inf)
    {
        for(i = 1; i <= m; i++)
            for(j = 1; j <= n;j++)
            printf("%d%c", rec[i][j], j == n ? '\n' : ' ');
    }
    else printf("IMPOSSIBLE\n");
    return 0;

}