[BZOJ 1647][USACO 2007 Open] Fliptile 翻格子游戏

1647: [Usaco2007 Open]Fliptile 翻格子游戏

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 702  Solved: 281
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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".

 约翰知道，那些高智力又快乐的奶牛产奶量特别高．所以他做了一个翻瓦片的益智游戏来娱乐奶牛．在一个M×N(1≤M，N≤15)的骨架上，每一个格子里都有一个可以翻转的瓦片．瓦片的一面是黑色的，而另一面是白色的．对一个瓦片进行翻转，可以使黑变白，也可以使白变黑．然而，奶牛们的蹄子是如此的巨大而且笨拙，所以她们翻转一个瓦片的时候，与之有公共边的相邻瓦片也都被翻转了．那么，这些奶牛们最少需要多少次翻转，使所有的瓦片都变成白面向上呢？如杲可以做到，输出字典序最小的结果（将结果当成字符串处理）．如果不能做到，输出“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行输入M和N，之后M行N列，输入游戏开始时的瓦片状态．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.

    输出M行，每行N个用空格隔开的整数，表示对应的格子进行了多少次翻转．

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.

第一眼看到数据范围可能很多人会想到状压 $DP$ , 然而看起来似乎是最大 $15 \times 15 = 225$ 的状压量根本无法枚举.

但是实际上我们可以推知, 根据上一行的黑白情况我们可以推知下一行的翻转情况. 因为在上一行已经确定的情况下下一行必须保证在黑色瓦片正下方进行翻转. 所以我们的所有情况都可以从第一行的黑白情况推知. 这时枚举量就从 $2^{k^2}$ 降到了 $2^k$ . 然后我们根据第一行枚举得出的黑白情况计算该种情况是否可以得出解. 如果得出解的话更新答案, 无解忽略即可.

枚举中途可以选择进行剪枝, 保存最低翻转数, 当已使用的翻转数大于这个值时停止继续求值.

参考代码

GitHub

 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <iostream>
 #include <algorithm>

 const int MAXN=;

 int n;
 int m;
 int sum;
 int flip;
 int minfl;
 int black;
 bool ans[MAXN][MAXN];
 bool raw[MAXN][MAXN];
 bool tmp[MAXN][MAXN];
 bool data[MAXN][MAXN];

 void Change(int,int);
 void Initialize();
 bool Check(int);
 bool Compare();

 int main(){
     freopen("fliptile.in","r",stdin);
     freopen("fliptile.out","w",stdout);
     bool solve=false;
     Initialize();
     for(int i=;i<(<<n);i++){
         memset(tmp,,sizeof(tmp));
         memcpy(data,raw,sizeof(raw));
         flip=;
         black=sum;
         if(Check(i)){
             memcpy(ans,tmp,sizeof(tmp));
             minfl=flip;
             solve=true;
         }
     }
     if(solve){
         for(int i=;i<=n;i++){
             for(int j=;j<=m;j++){
                 printf("%d ",ans[i][j]?:);
             }
             putchar('\n');
         }
     }
     else
         puts("IMPOSSIBLE");
     return ;
 }

 inline void Change(int x,int y){
     tmp[x][y]=true;
     data[x][y]=!data[x][y];
     black+=data[x][y]?:-;
     if(x>){
         data[x-][y]=!data[x-][y];
         black+=data[x-][y]?:-;
     }
     if(x<n){
         data[x+][y]=!data[x+][y];
         black+=data[x+][y]?:-;
     }
     if(y>){
         data[x][y-]=!data[x][y-];
         black+=data[x][y-]?:-;
     }
     if(y<m){
         data[x][y+]=!data[x][y+];
         black+=data[x][y+]?:-;
     }
 }

 void Initialize(){
     scanf("%d%d",&n,&m);
     int tmp=;
     minfl=0x7FFFFFFF;
     for(int i=;i<=n;i++){
         for(int j=;j<=m;j++){
             scanf("%d",&tmp);
             raw[i][j]=tmp;
             sum+=tmp;
         }
     }
 }

 bool Check(int x){
     for(int i=;i<n;i++){
         if(x&(<<i)){
             Change(,i+);
             flip++;
             if(flip>=minfl)
                 return false;
         }
     }
     for(int i=;i<=n;i++){
         for(int j=;j<=m;j++){
             if(data[i-][j]){
                 Change(i,j);
                 flip++;
                 if(flip>=minfl)
                     return false;
             }
         }
     }
     if(black>)
         return false;
     else
         return true;
 }

Backup

以及日常凸包图包