字典 esc and lin sds board side ant opened

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 x 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".

約翰知道，那些高智力又快樂的奶牛產奶量特別高．所以他做了一個翻瓦片的益智遊戲來娛樂奶牛．在一個M×N(1≤M，N≤15)的骨架上，每一個格子裏都有一個可以翻轉的瓦片．瓦片的一面是黑色的，而另一面是白色的．對一個瓦片進行翻轉，可以使黑變白，也可以使白變黑．然而，奶牛們的蹄子是如此的巨大而且笨拙，所以她們翻轉一個瓦片的時候，與之有公共邊的相鄰瓦片也都被翻轉了．那麽，這些奶牛們最少需要多少次翻轉，使所有的瓦片都變成白面向上呢？如杲可以做到，輸出字典序最小的結果（將結果當成字符串處理）．如果不能做到，輸出“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

第1行輸入M和N，之後M行N列，輸入遊戲開始時的瓦片狀態．0表示白面向上，1表示黑面向上．

Output

* Lines 1..M: Each line contains N space-separated integers, each specifying how many times to flip that particular location.

輸出M行，每行N個用空格隔開的整數，表示對應的格子進行了多少次翻轉．

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

OUTPUT DETAILS:

After flipping at row 2 column 1, the board will look like:
0 0 0 1
1 0 1 0
1 1 1 0
1 0 0 1

After flipping at row 2 column 4, the board will look like:
0 0 0 0
1 0 0 1
1 1 1 1
1 0 0 1

After flipping at row 3 column 1, the board will look like:
0 0 0 0
0 0 0 1
0 0 1 1
0 0 0 1

After flipping at row 3 column 4, the board will look like:
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0

Another solution might be:
0 1 1 0
0 0 0 0
0 0 0 0
0 1 1 0
but this solution is lexicographically higher than the solution above. solution 開始的時候打了dfs，看了題解之後立刻跪下....... 首先每個點只能翻1或0次，因為翻兩次相當於翻過來又翻過去，與0次沒有區別(類比異或) 有一個神奇但是不難看出(然而我沒看出來)的性質： 當第一行每個點翻不翻確定時，整個矩陣就確定了 因為目標是全0.....

  1 #include<cstdio>
  2 #include<iostream>
  3 #include<cstring>
  4 #define mem(a,b) memset(a,b,sizeof(a))
  5 #define cop(a,b) memcpy(a,b,sizeof(a))
  6 using namespace std;
  7 
  8 int n,m;
  9 int a[17][17],now[17][17];
 10 int b[17][17],ans[17][17];
 11 int maxp;
 12 int judge;
 13 
 14 void bian(int x,int y)
 15 {
 16     b[x][y]^=1;
 17     if(x>1)b[x-1][y]^=1;
 18     if(x<n)b[x+1][y]^=1;
 19     if(y>1)b[x][y-1]^=1;
 20     if(y<m)b[x][y+1]^=1;
 21 }
 22 
 23 int comp()
 24 {
 25     int numnow=0,numans=0;
 26     for(int i=1;i<=n;++i)
 27       for(int j=1;j<=m;++j)
 28       {
 29             numnow+=now[i][j];
 30             numans+=ans[i][j];
 31         }
 32     if(numnow!=numans)
 33       return numnow<numans;
 34     for(int i=1;i<=n;++i)
 35       for(int j=1;j<=n;++j)
 36         if(now[i][j]!=ans[i][j])
 37           return now[i][j]<ans[i][j];
 38 }
 39 
 40 void jilu()
 41 {
 42     //printf("sdadsdsa\n");
 43     if(!judge)
 44     {
 45         judge=1;
 46         cop(ans,now);
 47         return ;
 48     }
 49     if(comp())
 50       cop(ans,now);
 51 }
 52 
 53 int main(){
 54     //freopen("fliptile6.in","r",stdin);
 55     //freopen("fliptile.out","w",stdout);
 56     scanf("%d%d",&n,&m);
 57     maxp=(1<<m)-1;
 58     for(int i=1;i<=n;++i)
 59       for(int j=1;j<=m;++j)
 60         scanf("%d",&a[i][j]);
 61     
 62     for(int k=0;k<=maxp;++k)
 63     {
 64         cop(b,a);
 65         for(int i=1;i<=m;++i)
 66         {
 67             if(k&(1<<(i-1)))
 68             {
 69               now[1][i]=1;
 70               bian(1,i);
 71             }
 72             else
 73               now[1][i]=0;
 74         }
 75         for(int i=2;i<=n;++i)
 76           for(int j=1;j<=m;++j)
 77           {
 78             if(b[i-1][j])
 79             {
 80                     now[i][j]=1;
 81                     bian(i,j);
 82                 }
 83                 else
 84                   now[i][j]=0;
 85             }
 86         
 87         int flag=0;
 88         for(int i=1;i<=m;++i)
 89           if(b[n][i])
 90             flag=1;
 91         if(!flag)
 92           jilu();
 93     }
 94     
 95     if(judge)
 96       for(int i=1;i<=n;++i)
 97       {
 98             for(int j=1;j<=m;++j)
 99               printf("%d ",ans[i][j]);
100             printf("\n");
101         }
102     else
103       printf("IMPOSSIBLE");
104     //while(1);
105     return 0;
106 }

code

[Usaco2007 Open]Fliptile 翻格子遊戲