标签：poj3279
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 3487 | Accepted: 1351 |
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
Output
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
/*
** Problem: POJ3279
** Status: Accepted
** Running time: 407ms
** Complier: C++
** Author: Changmu
**
** 题意：有一个M*N的棋盘，每一个格子只有两种状态0或1，每次可以选择一个
** 格子执行翻转操作，并且与该格子相邻的4个格子都会被翻转，求将所有格子
** 都翻转成0所需要的最小操作数，若有多种方案，输出字典序最小的方案数。
**
** 题解：枚举第一行的所有翻转状态，在此基础上深搜剩余行的翻转状态。
*/
#include <stdio.h>
#include <string.h>
#define maxn 16
#define inf 0x3f3f3f3f
int cow[maxn][maxn], M, N; // 原始数据 M rows
int flip[maxn][maxn]; // 保存中间结果
int opt[maxn][maxn]; // 保存最优结果
const int mov[][2] = {0, 0, 0, -1, 0, 1, -1, 0}; // 中左右上
void getMap() {
int i, j;
for(i = 0; i < M; ++i)
for(j = 0; j < N; ++j)
scanf("%d", &cow[i][j]);
}
int getColo(int x, int y) {
int c = cow[x][y], xa, ya, i;
for(i = 0; i < 4; ++i) {
xa = x + mov[i][0];
ya = y + mov[i][1];
if(xa >= 0 && xa < M && ya >= 0 && ya < N) {
c += flip[xa][ya];
}
}
return c & 1;
}
int DFS(int k) {
int i, j, sum;
if(k == M - 1) { // 最后一行不再翻转，判断是否都0
for(i = 0; i < N; ++i)
if(getColo(k, i)) break;
if(i != N) return -1;
// 统计翻转次数
for(i = sum = 0; i < M; ++i)
for(j = 0; j < N; ++j)
if(flip[i][j]) ++sum;
return sum;
}
for(j = 0; j < N; ++j)
if(getColo(k, j)) flip[k + 1][j] = 1;
return DFS(k + 1);
}
void solve() {
int i, j, maxCase = 1 << N, ret = inf, num, tmp;
for(i = 0; i < maxCase; ++i) {
memset(flip, 0, sizeof(flip));
for(j = N - 1, tmp = i; j >= 0; --j, tmp >>= 1) {
flip[0][j] = tmp & 1;
}
num = DFS(0);
if(num != -1 && num < ret) {
ret = num;
memcpy(opt, flip, sizeof(flip));
}
}
if(ret == inf) printf("IMPOSSIBLE\n");
else {
for(i = 0; i < M; ++i)
for(j = 0; j < N; ++j)
if(j == N - 1) printf("%d\n", opt[i][j]);
else printf("%d ", opt[i][j]);
}
}
int main() {
// freopen("stdin.txt", "r", stdin);
while(scanf("%d%d", &M, &N) == 2) {
getMap();
solve();
}
return 0;
}
标签：poj3279
原文地址：http://blog.csdn.net/chang_mu/article/details/40895325
