Fliptile [反转问题][dfs]_flip过后net也反了-优快云博客

本文介绍了一种通过翻转网格中的方格来使所有方格颜色统一的问题，并提供了一个有效的解决方案。采用递归搜索的方式，从第一行开始尝试不同的翻转组合，最终找到最小翻转次数及翻转位置。

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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

解题报告

只要确定第一行，下面的行就都确定了，枚举第一行即可，复杂度o(N*M * 2^N)

//300 MS

#include<stdio.h>
#include<string.h>
#define MAX_N 15
int map[MAX_N][MAX_N];
int tmp_map[MAX_N][MAX_N];
bool vis[MAX_N][MAX_N];
bool tmp_vis[MAX_N][MAX_N];
bool bests[MAX_N][MAX_N];
int M,N,best;
int ox[]={0,-1,0,0,1};
int oy[]={-1,0,0,1,0};

void flag(int Y,int X,int (*P_map)[MAX_N],bool (*P_vis)[MAX_N]){
    for(int i=0;i<5;i++){
        int x=X+ox[i];
        int y=Y+oy[i];
        if(0<=x&&x<N&&0<=y&&y<M)
            P_map[y][x]++;
    }
    P_vis[Y][X]=!P_vis[Y][X];
}

void go(int cnt){
    //初始化
    memset(tmp_vis,0,sizeof(tmp_vis));
    for(int j=0;j<M;j++)
        for(int k=0;k<N;k++)
            tmp_map[j][k]=map[j][k];
    for(int i=0;i<N;i++) tmp_vis[0][i]=vis[0][i];

    for(int j=1;j<M;j++)
        for(int k=0;k<N;k++){
            if(tmp_map[j-1][k]&1)
                flag(j,k,tmp_map,tmp_vis),cnt++;
            if(cnt>=best) return ;
        }
    for(int i=0;i<N;i++)
        if(tmp_map[M-1][i]&1)
            return ;
    if(cnt<best){
        for(int j=0;j<M;j++)
            for(int k=0;k<N;k++)
                bests[j][k]=tmp_vis[j][k];
        best=cnt;
    }
}

void dfs(int x,int cnt){
    if(x==N){
        go(cnt);
        return ;
    }

    dfs(x+1,cnt);

    flag(0,x,map,vis);
    dfs(x+1,cnt+1);
    flag(0,x,map,vis);
}

int main()
{
    while(~scanf("%d%d",&M,&N)){
        memset(vis,0,sizeof(vis));
        for(int j=0;j<M;j++)
            for(int k=0;k<N;k++)
                scanf("%d",&map[j][k]);
        best=0x3f3f3f3f;dfs(0,0);
        if(best!=0x3f3f3f3f)
            for(int j=0;j<M;j++){
                for(int k=0;k<N;k++)
                    printf(k?" %d":"%d",bests[j][k]);
                putchar('\n');
            }
        else puts("IMPOSSIBLE");
    }
    return 0;
}