POJ 3279 枚举（思维）

Fliptile

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10931		Accepted: 4029

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

思路：1.先枚举第一行的翻转情况，然后相邻的下一行的翻转情况就已经确定，因为这时只有下一行的翻转能改变当前行的状态，这样每个状态只由第一行确定。

2.要保证字典序最小，只需使第一行的字典序最小。

代码：

#include "cstdio"
#include "stdlib.h"
#include "iostream"
#include "algorithm"
#include "string"
#include "cstring"
#include "queue"
#include "cmath"
#include "vector"
#include "map"
#include "set"
#define mj
#define db double
#define ll long long
using namespace std;
const int inf = 0x3f3f3f3f;
const int N = 17;
const int dx[] = {-1, 0, 1, 0, 0};
const int dy[] = { 0,-1, 0, 1, 0};

int s[N][N], st[N][N], tmp[N][N], rec[N][N];
int n, m, ans;
void flip(int x, int y) {
    tmp[x][y] = 1;
    int nx, ny;
    for(int i = 0; i < 5; i++) {
        nx = x+dx[i];
        ny = y+dy[i];
        st[nx][ny] = !st[nx][ny];
    }
}
bool ok() {
    for(int j = 1; j <= m; j++) {
        if(st[n][j]) return 0;
    }
    return 1;
}
void f(int t) {
    memcpy(st, s, sizeof(s));
    memset(tmp, 0, sizeof(tmp));
    int cnt = 0;
    for(int j = 0; j < m; j++) {
        if((t>>j) & 1) {//从第一行的二进制第一位开始枚举翻转状态。
            flip(1, j+1);
            cnt++;
        }
    }
    for(int i = 2; i <= n; i++) {
        for(int j = 1; j <= m; j++) {
            if(st[i-1][j] == 1) {
                flip(i, j);
                cnt++;
            }
        }
    }
    if(ok() && cnt < ans) {
        ans = cnt;
        memcpy(rec, tmp, sizeof(tmp));
    }
}

int main() {
    while(scanf("%d%d", &n, &m) != EOF) {
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                scanf("%d", &s[i][j]);
            }
        }
        ans = inf;
        int end = 1 << m;
        for(int t = 0; t < end; t++)  f(t);
        if(ans == inf) puts("IMPOSSIBLE");
        else {
            for(int i = 1; i <= n; i++) {
                printf("%d", rec[i][1]);
                for(int j = 2; j <= m; j++)
                    printf(" %d", rec[i][j]);
                puts("");
            }
        }
    }
    return 0;
}