Fliptile

  　　　　　　　　　　　　　开关题   尺度法

　　　　　 Fliptile

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3394		Accepted: 1299

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

 1 #include"iostream"
 2 #include"cstdio"
 3 #include"cstring"
 4 #include"algorithm"
 5 using namespace std;
 6 const int ms=16;
 7 int dx[5]={-1,0,0,0,1};
 8 int dy[5]={0,-1,0,1,0};
 9 int M,N;
10 int tile[ms][ms];
11 int opt[ms][ms];//保存最优解
12 int flip[ms][ms];
13 int get(int x,int y)
14 {
15     int c=tile[x][y];
16     for(int d=0;d<5;d++)
17     {
18         int x2=x+dx[d];
19         int y2=y+dy[d];
20         if(x2>=0&&x2<M&&y2>=0&&y2<N)
21             c+=flip[x2][y2];
22     }
23     return c%2;
24 } 
25 //  求出第一行确定的情况下，最小的操作次数
26 //  第一行确定了所有行都确定了。 
27 int calc()
28 {
29     for(int i=1;i<M;i++)
30     {
31         for(int j=0;j<N;j++)
32         {
33             if(get(i-1,j))
34                 flip[i][j]=1;
35         }
36     }
37     for(int j=0;j<N;j++)
38         if(get(M-1,j))
39             return -1;
40     int res=0;
41     for(int i=0;i<M;i++)
42         for(int j=0;j<N;j++)
43             res+=flip[i][j];
44     return res;
45 } 
46 void solve()
47 {
48     int res=-1;
49     //按照字典序尝试第一行的所有可能性
50     for(int i=0;i<1<<N;i++)
51     {
52         memset(flip,0,sizeof(flip));
53         for(int j=0;j<N;j++)
54             flip[0][N-j-1]=i>>j&1;
55         int num=calc();
56         if(num>=0&&(res<0||res>num))
57         {
58             res=num;
59             memcpy(opt,flip,sizeof(flip));
60         }
61     } 
62     if(res<0)
63         printf("IMPOSSIBLE\n");
64     else
65         for(int i=0;i<M;i++)
66             for(int j=0;j<N;j++)
67                 printf("%d%c",opt[i][j],j+1==N?'\n':' ');
68     return ;
69 }
70 int main()
71 {
72     scanf("%d%d",&M,&N);
73     for(int i=0;i<M;i++)
74         for(int j=0;j<N;j++)
75         {
76             scanf("%d",&tile[i][j]);
77         }
78     solve();
79     return 0;
80 }

 

转载于:https://www.cnblogs.com/767355675hutaishi/p/3952241.html