POJ 3279 Fliptile

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本文介绍了一个名为Fliptile的游戏问题及其算法解决方案。该问题是让玩家在一个M×N的网格上翻转黑白瓷砖，目标是通过最少的操作使所有瓷砖白色面朝上。文章详细解释了如何使用二进制枚举来确定最佳的第一行翻转方案，并通过递推确定后续行的操作。

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来自<http://poj.org/problem?id=3279>

Fliptile

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 13959		Accepted: 5153

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

Sample Output

Source

USACO 2007 Open Silver

题意：n*m的方格中，将所有的1变为0，选择一个进行转变会引起周围四个一起转变。输出结果是n*m的矩阵，1代表对这个点进行转变，0代表不进行转变，问转变次数最少且第一行字典序最小的方案

题解：用数的二进制枚举第一行的方案，后面的可以递推出来，检查最后一行是否满足条件

#include<iostream>
#include<cstdio>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
typedef long long LL;
int n,m,len;
int map[20][20];
int vis[20][20];
int tmp[20][20];
int ans[20][20];
void cg(int i,int j){if(vis[i][j]==1) vis[i][j]=0;else vis[i][j]=1;}
void change(int i,int j){
	cg(i-1,j);cg(i+1,j);
	cg(i,j+1);cg(i,j-1);
	cg(i,j);
}
int main(){
    int cnt=0,mi=(int)1e9;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            scanf("%d",&map[i][j]);
    len=1<<m;
    for(int i=0;i<len;i++){
    	memset(tmp,0,sizeof(tmp));
    	memcpy(vis,map,sizeof(map));
    	cnt=0;
        for(int j=1;j<=m;j++){
            if(i&(1<<(m-j))) tmp[1][j]=1;
            if(tmp[1][j]==1){cnt++;change(1,j);}
        }
        for(int j=2;j<=n;j++){
            for(int k=1;k<=m;k++){
                if(vis[j-1][k]==1){
                    cnt++;
                    tmp[j][k]=1;
                    change(j,k);
                }
            }
        }
        int flag=1;
        for(int k=1;k<=m;k++){
            if(vis[n][k]==1){
                flag=0;break;
            }
        }
        if(flag&&cnt<mi){
            memcpy(ans,tmp,sizeof(tmp));
            mi=cnt;
        }
    }
    if(mi<(int)1e9){
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                printf("%d ",ans[i][j]);
            }
            printf("\n");
        }
    }
    else{
        printf("IMPOSSIBLE\n");
    }
}