1647: [Usaco2007 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: 423  Solved: 173
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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.

HINT

 

Source

Silver

 

题解：没记错的话，这个是当年囧神（囧神=JSZKC，省选前夕orz一下攒攒RP）出的NOIP模拟题里面的$T_3$，当时的我只知道 $O({2}^{NM})$ 的纯暴力枚举，但事实上不用这样——

其实还是枚举，但实际上只要 $O({2}^{N} )$ 的枚举即可，看到 $N$ ，显然就是枚举第一行啦，事实上第一行的决策将直接决定下一行的决策，然后下一行影响下一行，也就是说实际上第一行的决定决定了全局的决策，然后根据第一行二进制穷举出来的进行模拟，然后判断此方案是否可行，然后打擂台记录下来

有人可能会在比对过程中再弄个字典序比较，因为题目中说要字典序最小，但是还有一个细节你似乎忽略了——我们二进制穷举从小数字到大数字本身就符合字典序上升，对于一个第一行决策情况，最多只有一种合法解，所以完全不必比较，直接“先入为主”即可

/**************************************************************
    Problem: 1647
    User: HansBug
    Language: Pascal
    Result: Accepted
    Time:500 ms
    Memory:232 kb
****************************************************************/
 
var
   i,j,k,l,m,n,ans:longint;
   a,c,e,f:array[0..20,0..20] of longint;
   b:array[0..20] of longint;
begin
     readln(n,m);
     for i:=1 to n do
         begin
              for j:=1 to m do read(a[i,j]);
              readln;
         end;
     fillchar(b,sizeof(b),0);ans:=maxlongint;
     while b[0]=0 do
           begin
                for i:=1 to m do c[0,i]:=b[i];
                for i:=1 to n do
                    for j:=1 to m do c[i,j]:=a[i,j];
                fillchar(e,sizeof(e),0);k:=0;
                for i:=1 to n do
                    begin
                         for j:=1 to m do
                             if c[i-1,j]=1 then
                                begin
                                     e[i,j]:=1;inc(k);
                                     c[i,j]:=1-c[i,j];
                                     c[i,j-1]:=1-c[i,j-1];
                                     c[i,j+1]:=1-c[i,j+1];
                                     c[i-1,j]:=1-c[i-1,j];
                                     c[i+1,j]:=1-c[i+1,j];
                                end;
                    end;
                l:=0;
                for i:=1 to m do inc(l,c[n,i]);
                if l=0 then
                   begin
                        if k<ans then
                           begin
                                ans:=k;
                                for i:=1 to n do
                                    for j:=1 to m do
                                        f[i,j]:=e[i,j];
                           end;
                   end;
                i:=m;
                while b[i]=1 do
                      begin
                           b[i]:=0;
                           dec(i);
                      end;
                b[i]:=1;
           end;
     if ans=maxlongint then
        begin
             writeln('IMPOSSIBLE');
             halt;
        end;
     for i:=1 to n do
         for j:=1 to m do
             if j<m then write(f[i,j],' ') else writeln(f[i,j]);
     readln;
end.