题目链接：http://poj.org/problem?id=3279
| Time Limit:2000MS | Memory Limit:65536K | |
| Total Submissions:11771 | Accepted:4360 |
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 anM×Ngrid (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
Lines 2..M+1: Linei+1 describes the colors (left to right) of row i of the grid withNspace-separated integers which are 1 for black and 0 for white
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
题解：
1.枚举第一行的所有状态。
2.从第二行开始，如果当前瓦片的头上瓦片为1，那么需要翻转当前瓦片。因为在这一行之内，只有当前瓦片能够使头上的瓦片变为0，然后一直递推到最后一行。
3.分析：一开始想到方法是枚举，但是数据太大，计算机要算多少年真的不好估计。先假设枚举法是可行的， 相对于枚举法， 上述方法是有明显的优势的：上述方法从第二行开始，每翻转一块瓦片都是有针对性的、有目的性的；而枚举法则是盲目地翻转，如：假如当前瓦片为0，却还是硬要翻转该瓦片，那显然是行不通的。所以上述方法相对来说更“智能”。
4.注意：s&(1<<i)的值要么等于0，要么等于(1<<i)；而不是0或1。应注意！！
5.此题的弱化版，可使用枚举法。POJ1753：http://www.cnblogs.com/DOLFAMINGO/p/7538788.html
代码如下：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <vector> 7 #include <queue> 8 #include <stack> 9 #include <map> 10 #include <string> 11 #include <set> 12 #define ms(a,b) memset((a),(b),sizeof((a))) 13 using namespacestd; 14 typedef long longLL; 15 const int INF =2e9; 16 const LL LNF =9e18; 17 const int MOD = 1e9+7; 18 const int MAXN = 15+10; 19 20 intM[MAXN][MAXN], subM[MAXN][MAXN]; 21 intop[MAXN][MAXN], ans_op[MAXN][MAXN]; 22 intn, m; 23 24 void press(int x, inty) 25 { 26 op[x][y] = 1; 27 subM[x][y] = !subM[x][y]; 28 if(x-1>=0) subM[x-1][y] = !subM[x-1][y]; 29 if(x+1<n) subM[x+1][y] = !subM[x+1][y]; 30 if(y-1>=0) subM[x][y-1] = !subM[x][y-1]; 31 if(y+1<m) subM[x][y+1] = !subM[x][y+1]; 32 } 33 34 boolallZero() 35 { 36 for(int i = 0; i<n; i++) 37 for(int j = 0; j<m; j++) 38 if(subM[i][j]) return false; 39 return true; 40 } 41 42 voidsolve() 43 { 44 int ans_step =INF; 45 for(int s = 0; s<(1<<m); s++) 46 { 47 ms(op, 0); 48 memcpy(subM, M, sizeof(subM)); 49 int step = 0; 50 for(int i = 0; i<m; i++) 51 if(s&(1<<i)) press(0, i), step++; 52 53 for(int i = 1; i<n; i++) 54 for(int j = 0; j<m; j++) 55 if(subM[i-1][j]) press(i, j), step++; 56 57 if(allZero() && ans_step>step ) 58 { 59 ans_step =step; 60 memcpy(ans_op, op, sizeof(ans_op)); 61 } 62 } 63 if(ans_step==INF) 64 puts("IMPOSSIBLE"); 65 else 66 { 67 for(int i = 0; i<n; i++){ 68 for(int j = 0; j<m; j++) 69 printf("%d ",ans_op[i][j]); 70 putchar(''); 71 } 72 } 73 } 74 75 intmain() 76 { 77 while(scanf("%d%d",&n,&m)!=EOF) 78 { 79 for(int i = 0; i<n; i++) 80 for(int j = 0; j<m; j++) 81 scanf("%d",&M[i][j]); 82 83 solve(); 84 } 85 return 0; 86 }