[BZOJ2462] [BeiJing2011]矩阵模板(二维Hash)

传送门

 

二维哈希即可。

注意质数选的大一些,不然会超时。

还有插入的时候不判重居然比判重要快。。

 

——代码

1 #include <cstdio>
2 int main()
3 {
4     int i = 10;
5     while(i--) puts("1");
6     return 0;
7 }
View Code

 

O不,错了,是这个。

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #define UI unsigned int
 5 
 6 const int p = 1000007, MAXN = 1001;
 7 int n, m, a, b, q, cnt, head[p], next[MAXN * MAXN];
 8 UI h, sum[MAXN][MAXN], base1[MAXN], base2[MAXN], to[MAXN * MAXN];
 9 
10 inline void insert(UI x)
11 {
12     int i, a = x % p;
13     for(i = head[a]; i ^ -1; i = next[i])
14         if(!(to[i] ^ x))
15             return;
16     to[cnt] = x;
17     next[cnt] = head[a];
18     head[a] = cnt++;
19 }
20 
21 inline bool find(UI x)
22 {
23     int i, a = x % p;
24     for(i = head[a]; i ^ -1; i = next[i])
25         if(!(to[i] ^ x))
26             return 1;
27     return 0;
28 }
29 
30 int main()
31 {
32     int i, j;
33     scanf("%d %d %d %d", &n, &m, &a, &b);
34     base1[0] = base2[0] = 1;
35     memset(head, -1, sizeof(head));
36     for(i = 1; i <= n; i++) base1[i] = base1[i - 1] * 19260817;
37     for(i = 1; i <= m; i++) base2[i] = base2[i - 1] * 20011001;
38     for(i = 1; i <= n; i++)
39         for(j = 1; j <= m; j++)
40             scanf("%1d", &sum[i][j]);
41     for(i = 1; i <= n; i++)
42         for(j = 1; j <= m; j++)
43             sum[i][j] += sum[i - 1][j] * 19260817;
44     for(i = 1; i <= n; i++)
45         for(j = 1; j <= m; j++)
46             sum[i][j] += sum[i][j - 1] * 20011001;
47     for(i = a; i <= n; i++)
48         for(j = b; j <= m; j++)
49         {
50             h = sum[i][j];
51             h -= sum[i - a][j] * base1[a];
52             h -= sum[i][j - b] * base2[b];
53             h += sum[i - a][j - b] * base1[a] * base2[b];
54             insert(h);
55         }
56     scanf("%d", &q);
57     while(q--)
58     {
59         for(i = 1; i <= a; i++)
60             for(j = 1; j <= b; j++)
61                 scanf("%1d", &sum[i][j]);
62         for(i = 1; i <= a; i++)
63             for(j = 1; j <= b; j++)
64                 sum[i][j] += sum[i - 1][j] * 19260817;
65         for(i = 1; i <= a; i++)
66             for(j = 1; j <= b; j++)
67                 sum[i][j] += sum[i][j - 1] * 20011001;
68         printf("%d\n", find(sum[a][b]));
69     }
70     return 0;
71 }
View Code

 

posted @ 2017-05-16 16:15  zht467  阅读(...)  评论(...编辑  收藏