TyvjP1266 费解的开关

 1 #include <bits/stdc++.h>
 2 
 3 inline void read(int &x)
 4 {
 5     x = 0;char ch = getchar();
 6     char c = ch;
 7     while(ch > ‘9‘ || ch < ‘0‘)c = ch, ch = getchar();
 8     while(ch <= ‘9‘ && ch >= ‘0‘)x = x * 10 + ch - ‘0‘, ch = getchar();
 9     if(c == ‘-‘)x = -x;
10 }
11 inline int min(int a, int b){return a > b ? b : a;}
12 
13 const int INF = 0x3f3f3f3f; 
14 const int MAXN = 50 + 10;
15 
16 int g[MAXN][MAXN],n,ans,tmp[MAXN][MAXN];
17 
18 void change(int x, int y)
19 {
20     tmp[x][y] ^= 1;
21     if(x - 1)
22         tmp[x - 1][y] ^= 1;
23     if(y - 1)
24         tmp[x][y - 1] ^= 1;
25     if(y + 1 <= n)
26         tmp[x][y + 1] ^= 1;
27     if(x + 1 <= n)
28         tmp[x + 1][y] ^= 1;
29 }
30 
31 int t;
32 char s[MAXN];
33 
34 int main()
35 {
36     read(t);
37     n = 5;
38     for(;t;--t)
39     {
40         memset(g, 0, sizeof(g));
41         memset(tmp, 0, sizeof(tmp));
42         ans = INF;
43         for(int i = 1;i <= n;++ i)
44         {
45             scanf("%s", s + 1);
46             for(int j = 1;j <= n;++ j)
47                 g[i][j] = (s[j] - ‘0‘) ^ 1;
48         }
49         for(register int s = 0;s < (1 << n);++ s)
50         {
51             for(int i = 1;i <= n;++ i)
52                 for(int j = 1;j <= n;++ j)
53                     tmp[i][j] = g[i][j];
54             bool ok = true;
55             register int tmp = s;
56             register int k = 1;
57             int cnt = 0;
58             while(tmp)
59             {
60                 if(tmp & 1)
61                     change(1, k),++ cnt;
62                 ++ k;tmp >>= 1;
63             }
64             for(register int j = 2;j <= n;++ j)
65                 for(k = 1;k <= n;++ k)
66                     if(::tmp[j - 1][k])
67                         change(j, k), ++ cnt;
68             for(register int j = 1;j <= n;++ j)
69                 if(::tmp[n][j])
70                     ok = false;
71             if(ok)
72                 ans = min(ans, cnt);
73         }
74         if(ans > 6)
75             printf("-1\n");
76         else printf("%d\n", ans);
77     }
78     return 0;
79 }