# Painter's Problem

 Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 5378 Accepted: 2601

Description

There is a square wall which is made of n*n small square bricks. Some bricks are white while some bricks are yellow. Bob is a painter and he wants to paint all the bricks yellow. But there is something wrong with Bob's brush. Once he uses this brush to paint brick (i, j), the bricks at (i, j), (i-1, j), (i+1, j), (i, j-1) and (i, j+1) all change their color. Your task is to find the minimum number of bricks Bob should paint in order to make all the bricks yellow.

Input

The first line contains a single integer t (1 <= t <= 20) that indicates the number of test cases. Then follow the t cases. Each test case begins with a line contains an integer n (1 <= n <= 15), representing the size of wall. The next n lines represent the original wall. Each line contains n characters. The j-th character of the i-th line figures out the color of brick at position (i, j). We use a 'w' to express a white brick while a 'y' to express a yellow brick.

Output

For each case, output a line contains the minimum number of bricks Bob should paint. If Bob can't paint all the bricks yellow, print 'inf'.

Sample Input

2
3
yyy
yyy
yyy
5
wwwww
wwwww
wwwww
wwwww
wwwww


Sample Output

0

15

构造矩阵高斯消元后可以得到一组解，但是题目中要求的是求出最小染色次数。所以要对其中不确定的方案进行枚举。

 1 #include<iostream>
2 #include<cstdio>
3 #include<cstdlib>
4 #include<cmath>
5 #include<algorithm>
6 #include<cstring>
7 #include<queue>
8 #include<vector>
9 using namespace std;
10 int T,N,ANS;
11 int a[300][300];
12 bool gauss(){
13     int now=1;
14     for(int i=1;i<=N*N;i++){
15         int to=now;
16         while(to<=N*N&&a[to][i]==0) to++;
17         if(to>N*N) continue;
18         if(to!=now){
19             for(int j=1;j<=N*N+1;j++) swap(a[to][j],a[now][j]);
20         }
21         for(int j=1;j<=N*N;j++){
22             if(j!=now&&a[j][i]){
23                 for(int k=1;k<=N*N+1;k++){
24                     a[j][k]^=a[i][k];
25                 }
26             }
27         }
28         now++;
29     }
30     for(int i=now;i<=N*N;i++)
31         if(a[i][N*N+1]!=0) return false;
32     return true;
33 }
34
35 int v[300],cnt;
36 void dfs(int x){
37     if(cnt>=ANS) return ;//已经比目前的答案大了，没有必要再搜
38     if(x==0){
39         ANS=min(cnt,ANS);
40         return ;
41     }
42     if(a[x][x]!=0){
43         int num=a[x][N*N+1];//num表示第x块砖染色不染色
44         for(int i=x+1;i<=N*N;i++){
45             if(a[x][i]!=0) num=num^v[i];//已经枚举过的x+1~N*N中某块砖如果可以对x产生影响且已染色，就让num改变一次
46         }
47         v[x]=num;
48         if(num==1) cnt++;
49         dfs(x-1);
50         if(num==1) cnt--;
51     }
52     else{//枚举按或不按两种情况
53         v[x]=0; dfs(x-1);
54         v[x]=1; cnt++; dfs(x-1); cnt--;
55     }
56 }
57
58 int main(){
59     scanf("%d",&T);
60     while(T--){
61         memset(a,0,sizeof(a));
62         scanf("%d",&N);
63         for(int i=1;i<=N*N;i++){
64             a[i][i]=1;
65             if(i%N!=1) a[i][i-1]=1;
66             if(i%N!=0) a[i][i+1]=1;
67             if(i>=N+1) a[i][i-N]=1;
68             if(i<=N*(N-1)) a[i][i+N]=1;
69         }
70         for(int i=1;i<=N;i++){
71             char s[300];
72             scanf("%s",s+1);
73             for(int j=1;j<=N;j++){
74                 if(s[j]=='w') a[(i-1)*N+j][N*N+1]=1;
75             }
76         }
77         if(gauss()==false){
78             puts("inf");
79             continue;
80         }
81         ANS=1<<28;
82         dfs(N*N);
83         printf("%d\n",ANS);
84     }
85     return 0;
86 }

posted @ 2016-02-29 09:31  CXCXCXC  阅读(360)  评论(0编辑  收藏  举报