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 ",rfind(tail)-lfind(tail)+1); 83 } 84 }