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Fliptile

Time Limit: 2000MS Memory Limit: 65536K

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

比较难想的一题,但是这也是很经典的一题 经典游戏“点亮灯泡”就是这种 也就是按下一个开关,这盏灯及上下左右的灯会改变状态 因为一个开关按两次是没有意义的 所以对于每个开关只有两种状态 按或者不按 所以这一题就是要求输出 每一个开关按(1)或者不按(0)
  题目问是否能通过操作让所有的1变成0 如果不能则输出IMPOSSIBLE 如果有多种操作则输出需要按(1)的次数最少的 如果次数相等则输出结果字典序最小的那个 (如果有00111和11100则输出00111)   思路: 若枚举所有格子开或不开有2n*m个状态 所以在1≤M≤15,1≤N≤15的范围内肯定是要爆炸 所以思考能否优化   事实上,若第一行确定,则它的下一行的点法必然是确定的 因为对第一行有影响的只有第二行它对应的灯 所以若第一行灯没亮则第二行应该要按下开关使它点亮 以此类推可得所有的状态 如果最后一行在按完之后也全部点亮 则说明这种方案是成立的 所以可以枚举第一行的操作有2m个状态   代码: 为了方便表示出第一行灯的操作 可以使用二进制来表示是否操作
0000(0) 表示 否否否否 0001(1) 表示 否否否是 0010(2) 表示 否否是否 0011(3) 表示 否否是是
以此类推即可用循环把所有操作表示出来 共有2m次循环 对于每一个i可以进行且操作i&1为二进制最后一位数 而对于判断灯亮灭则可使用异或操作 因为我们有运算法则
0^1==1 1^1==0 0^0==0 1^0==1
也就是异或1则改变异或0则不变 这与灯是否被操作的情况是相同的   得到每种状态第一行的操作后 我们可以依次向下判断来更新答案
#include<stdio.h>
#include<string.h>
int map[20][20];
int temp[20][20];
int N,M;
void Pr(int x) {
    for(int i=0; i<M; i++) {
        map[0][i] ^= (x>>(M-1-i))&1;
        if((x>>(M-1-i))&1) {
            if(i>0)
                map[0][i-1] ^= 1;
            if(i<M-1)
                map[0][i+1] ^= 1;
            if(N>1)
                map[1][i] ^= 1;
        }
        printf("%d",(x>>(M-1-i))&1);
        printf("%c",i==M-1?'\n':' ');
    }
    for(int i=1; i<N; i++) {
        for(int j=0; j<M; j++) {
            printf("%d",map[i-1][j]);
            if(map[i-1][j]==1) {
                map[i-1][j] ^= 1;
                map[i][j] ^= 1;
                if(j>0)
                    map[i][j-1] ^= 1;
                if(j<M-1)
                    map[i][j+1] ^= 1;
                if(i<N-1)
                    map[i+1][j] ^= 1;
            }
            printf("%c",j==M-1?'\n':' ');
        }
    }
}
int main() {
    while(scanf("%d %d",&N,&M)!=EOF) {
        for(int i=0; i<N; i++)
            for(int j=0; j<M; j++)
                scanf("%d",&map[i][j]);
        int res=99999999,rest=-1;
        int T=(1<<M),max=T-1;
        while(T--) {
            memcpy(temp,map,sizeof(map));
            int t=T^max;
            int cnt=0;
            for(int i=0; i<M; i++) {
                temp[0][i] ^= (t>>(M-1-i))&1;
                if((t>>(M-1-i))&1) {
                    cnt++;
                    if(i>0)
                        temp[0][i-1] ^= 1;
                    if(i<M-1)
                        temp[0][i+1] ^= 1;
                    if(N>1)
                        temp[1][i] ^= 1;
                }
            }
            for(int i=1; i<N; i++) {
                for(int j=0; j<M; j++) {
                    if(temp[i-1][j]==1) {
                        cnt++;
                        temp[i-1][j] ^= 1;
                        temp[i][j] ^= 1;
                        if(j>0)
                            temp[i][j-1] ^= 1;
                        if(j<M-1)
                            temp[i][j+1] ^= 1;
                        if(i<N-1)
                            temp[i+1][j] ^= 1;
                    }
                }
            }
            int i;
            for(i=0; i<M; i++)
                if(temp[N-1][i]==1)
                    break;
            if(i==M&&cnt<res)
                res=cnt,rest=t;
        }
        if(res==99999999)
            printf("IMPOSSIBLE\n");
        else
            Pr(rest);
    }
    return 0;
}

题目地址:【POJ】[3279]Fliptile

查看原文:<a href=http://www.boiltask.com/blog/?p=1933>http://www.boiltask.com/blog/?p=1933</a>