UVA 11019

Problem H
Matrix Matcher
Input: Standard Input

Output: Standard Output

 

Given an N * M matrix, your task is to find the number of occurences of an X * Y pattern.

 

Input

The first line contains a single integer t(t ≤ 15), the number of test cases.

 

For each case, the first line contains two integers N and M (N, M ≤ 1000). The next N lines contain M characters each.

 

The next line contains two integers X and Y (X, Y ≤ 100). The next X lines contain Y characters each. 

 

Output

For each case, output a single integer in its own line, the number of occurrences.

 

Sample Input                               Output for Sample Input

2 
1 1
x 
1 1
y 
3 3
abc 
bcd
cde 
2 2
bc
cd

0

2

 

 

一个二维的hash,可以先对每一行hash,然后在对列hash

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<string.h>
 4 #include<vector>
 5 #include<algorithm>
 6 #include<map>
 7 using namespace std;
 8 typedef unsigned long long ull;
 9 #define mmax 100000+10
10 ull base,seed=131,ansx[1100][1100],ansy[1100][1100],aans[1001000];
11 int n,m;
12 char p[1100][1100],q[110][110];
13 int lfind(ull tail){
14     int l=0,r=n*m-1,mid,ans=-1;
15     while(l<=r){
16         mid=(l+r)>>1;
17         if(aans[mid]>=tail) {
18             if(aans[mid]==tail) ans=mid;
19             r=mid-1;
20         }
21         else l=mid+1;
22     }
23     return ans;
24 }
25 int rfind(ull tail){
26     int l=0,r=n*m-1,mid,ans=-1;
27     while(l<=r){
28         mid=(l+r)>>1;
29         if(aans[mid]<=tail) {
30             if(aans[mid]==tail) ans=mid;
31             l=mid+1;
32         }
33         else r=mid-1;
34     }
35     return ans;
36 }
37 int main(){
38     int t;cin>>t;
39     while(t--){
40         cin>>n>>m;getchar();
41         for(int i=0;i<n;i++)
42            scanf("%s",p[i]);
43       //     cout<<"------"<<endl;
44         int x,y;
45         cin>>x>>y;
46         for(int i=0;i<x;i++)
47             scanf("%s",q[i]);
48        // cout<<"------"<<endl;
49         if(x>n||y>m){
50             cout<<0<<endl;
51             continue;
52         }
53         memset(ansx,0,sizeof(ansx));
54         base=1;
55         for(int i=1;i<y;i++) base*=seed;
56         for(int i=0;i<n;i++){
57             for(int j=0;j<y;j++) ansx[i][0]=ansx[i][0]*seed+p[i][j];
58             for(int j=y;j<m;j++) ansx[i][j-y+1]=(ansx[i][j-y]-p[i][j-y]*base)*seed+p[i][j];
59         }
60         memset(ansy,0,sizeof(ansy));
61         base=1;
62         for(int i=1;i<x;i++) base*=seed;
63         for(int i=0;i<m;i++){
64             for(int j=0;j<x;j++) ansy[0][i]=ansy[0][i]*seed+ansx[j][i];
65             for(int j=x;j<n;j++) ansy[j-x+1][i]=(ansy[j-x][i]-ansx[j-x][i]*base)*seed+ansx[j][i];
66         }
67         int id=0;
68         for(int i=0;i<n;i++)
69             for(int j=0;j<m;j++)
70             aans[id++]=ansy[i][j];
71         sort(aans,aans+id);
72         ull tail=0;
73         for(int i=0;i<x;i++){
74             ull tmp=0;
75             for(int j=0;j<y;j++) tmp=tmp*seed+q[i][j];
76             tail=tail*seed+tmp;
77         }
78         if(lfind(tail)==-1){
79             cout<<0<<endl;
80             continue;
81         }
82         printf("%d\n",rfind(tail)-lfind(tail)+1);
83     }
84 }

 

posted on 2015-01-15 19:35  天凉了  阅读(225)  评论(0)  编辑  收藏  举报

导航