POJ 2965 The Pilots Brothers' refrigerator （枚举）

The Pilots Brothers' refrigerator

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 16216 Accepted: 6105 Special Judge

Description

The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a refrigerator.

There are 16 handles on the refrigerator door. Every handle can be in one of two states: open or closed. The refrigerator is open only when all handles are open. The handles are represented as a matrix 4х4. You can change the state of a handle in any location[i, j] (1 ≤ i, j ≤ 4). However, this also changes states of all handles in rowi and all handles in column j.

The task is to determine the minimum number of handle switching necessary to open the refrigerator.

Input

The input contains four lines. Each of the four lines contains four characters describing the initial state of appropriate handles. A symbol “+” means that the handle is in closed state, whereas the symbol “−” means “open”. At least one of the handles is initially closed.

Output

The first line of the input contains N – the minimum number of switching. The rest N lines describe switching sequence. Each of the lines contains a row number and a column number of the matrix separated by one or more spaces. If there are several solutions, you may give any one of them.

Sample Input

-+--

----

-+--

Sample Output

1 1

1 3

1 4

4 1

4 3

4 4

Source

Northeastern Europe 2004, Western Subregion

题意：

有一个4*4的正方形，每个格子上有一个符号，仅有“+”和“-”，要求将所有的符号都翻转成“-”。

规定：当翻转某个符号时，它所对应的行号和列号上符号都被翻转。

思路：

该题与POJ 1753 Flip Game类似，也有同个点翻转两次与未翻转效果相同的特点，因此只需考虑16点，每个点是否需要翻转的情况。枚举2 * （4 * 4）次即可。

但是该题压时间压得很紧，直接枚举全部它不让AC，后来马上加了个特判，就当输入时本来就全为的“-”的情况，就969MS刚好过～

/************************************************************************* 
    > File Name: POJ2965.cpp 
    > Author: BSlin 
    > Mail:   
    > Created Time: 2013年10月31日 星期四 15时47分23秒 
 ************************************************************************/  
  
#include <stdio.h>  
#define INF 20  
  
int map1[5][5],map2[5][5];  
  
int min(int a, int b) {  
    return a > b ? b : a;  
}  
  
void change(int x, int y) {  
    int i, j;  
    for(i=0; i<4; i++) {  
        map2[i][y] = 1 - map2[i][y];  
    }  
    for(j=0; j<4; j++) {  
        map2[x][j] = 1 - map2[x][j];  
    }  
    map2[x][y] = 1 - map2[x][y];  
}  
  
int  check(int s) {  
    int i, j, x, y;  
    int cnt;  
    for(i=0; i<4; i++) {  
        for(j=0; j<4; j++) {  
            map2[i][j] = map1[i][j];  
        }  
    }  
    cnt = 0;  
    for(i=0; i<16; i++) {  
        if(s & (1 << i)) {  
            cnt ++;  
            x = i / 4;  
            y = i % 4;  
            change(x,y);  
        }  
    }  
    for(i=0; i<4; i++) {  
        for(j=0; j<4; j++) {  
            if(map2[i][j] != 0) {  
                return INF;  
            }  
        }  
    }  
    return cnt;  
}  
  
int main() {  
    freopen("in.txt","r",stdin);  
  
    char ch;  
    int ans, i, j, now, num;  
    bool flag;  
    flag = false;  
    for(i=0; i<4; i++) {  
        for(j=0; j<4; j++) {  
            scanf("%c",&ch);  
            if(ch == '+') {  
                map1[i][j] = 1;  
                flag = true;  
            } else if(ch == '-') {  
                map1[i][j] = 0;  
            }  
        }  
        getchar();  
    }  
    if(flag == 0) {  
        printf("0\n");  
        return 0;  
    }  
    ans = INF;  
    for(i=0; i<65536; i++) {  
        now = check(i);  
        if(ans > now) {  
            ans = now;  
            num = i;  
        }  
    }  
    printf("%d\n",ans);  
    now = 32768;  
    for(i=15; i>=0; i--) {  
        if(num >= now) {  
            num -= now;  
            printf("%d %d\n",i/4+1, i%4+1);  
        }  
        now /= 2;  
    }  
}

补充：软件开发 , C++ ,