The Big Painting
Problem's Link: http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=88791
Mean:
给你两个由字符组成的矩阵,让你判断第一个矩阵在第二个矩阵中出现了多少次。
analyse:
典型的二维矩阵hash。
这题有两种做法:
第一种:横向hash,然后纵向KMP。时间复杂度是O(N*(N+M)).
第二种:二维hash。直接对两个矩阵做二维hash,然后判断大矩阵的子矩阵的hash值是否等于小矩阵的hash值,统计答案即可。
从实现上来说,二维hash更容易写,也更容易理解,但是我在比赛时忽略了一种情况:
1 2 2 4 xo xoxo oxox
正确答案是:3.
但如果我们在hash时,横向hash选取的种子值和纵向hash选取的种子值是相同的,就会出现错误,答案就变为了5。
这是因为本来答案只要统计横向的,但是由于横向和纵向的种子值相同,就会多统计进去两次纵向的。
如果题目说小矩形(或者大矩形)可以旋转匹配,那么种子值选取一样就对了。
Time complexity: 二维hash:O(N*M) hash+KMP:O(N*(N+M))
Source code:
二维hash:
/*
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-08-10-08.37
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#define LL long long
#define ULL unsigned long long
using namespace std;
const int MAXN=2010;
const ULL seed1 = 10000007; // line h
const ULL seed2 = 100000007;// row w
int hp,wp,hm,wm;
char G1[MAXN][MAXN],G2[MAXN][MAXN];
ULL Tab1[MAXN][MAXN],Tab2[MAXN][MAXN],goal;
void GetHash()
{
goal=0;
for(int i=0;i<hp;++i)
{
ULL a=0;
for(int j=0;j<wp;++j)
{
a=a*seed2+G1[i][j];
}
goal=goal*seed1+a;
}
}
int solve()
{
int ans=0;
ULL base1=1,base2=1;
for(int i=0;i<hp;++i) base1*=seed1;
for(int i=0;i<wp;++i) base2*=seed2;
for(int i=0;i<hm;++i) // line
{
ULL a=0;
for(int j=0;j<wp;++j) a=a*seed2+G2[i][j];
Tab1[i][wp-1]=a;
for(int j=wp;j<wm;++j)
{
Tab1[i][j]=Tab1[i][j-1]*seed2 + G2[i][j] -base2*G2[i][j-wp];
}
}
for(int i=wp-1;i<wm;++i)
{
ULL a=0;
for(int j=0;j<hp;++j) a=a*seed1+Tab1[j][i];
Tab2[hp-1][i]=a;
if(a==goal) ++ans;
for(int j=hp;j<hm;++j)
{
Tab2[j][i]=Tab2[j-1][i]*seed1 +Tab1[j][i] -Tab1[j-hp][i]*base1;
if(Tab2[j][i]==goal) ++ans;
}
}
return ans;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
while(~scanf("%d %d %d %d",&hp,&wp,&hm,&wm))
{
getchar();
for(int i=0;i<hp;++i) gets(G1[i]);
for(int i=0;i<hm;++i) gets(G2[i]);
GetHash();
printf("%d ",solve());
}
return 0;
}
/*
*/
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-08-10-08.37
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#define LL long long
#define ULL unsigned long long
using namespace std;
const int MAXN=2010;
const ULL seed1 = 10000007; // line h
const ULL seed2 = 100000007;// row w
int hp,wp,hm,wm;
char G1[MAXN][MAXN],G2[MAXN][MAXN];
ULL Tab1[MAXN][MAXN],Tab2[MAXN][MAXN],goal;
void GetHash()
{
goal=0;
for(int i=0;i<hp;++i)
{
ULL a=0;
for(int j=0;j<wp;++j)
{
a=a*seed2+G1[i][j];
}
goal=goal*seed1+a;
}
}
int solve()
{
int ans=0;
ULL base1=1,base2=1;
for(int i=0;i<hp;++i) base1*=seed1;
for(int i=0;i<wp;++i) base2*=seed2;
for(int i=0;i<hm;++i) // line
{
ULL a=0;
for(int j=0;j<wp;++j) a=a*seed2+G2[i][j];
Tab1[i][wp-1]=a;
for(int j=wp;j<wm;++j)
{
Tab1[i][j]=Tab1[i][j-1]*seed2 + G2[i][j] -base2*G2[i][j-wp];
}
}
for(int i=wp-1;i<wm;++i)
{
ULL a=0;
for(int j=0;j<hp;++j) a=a*seed1+Tab1[j][i];
Tab2[hp-1][i]=a;
if(a==goal) ++ans;
for(int j=hp;j<hm;++j)
{
Tab2[j][i]=Tab2[j-1][i]*seed1 +Tab1[j][i] -Tab1[j-hp][i]*base1;
if(Tab2[j][i]==goal) ++ans;
}
}
return ans;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
while(~scanf("%d %d %d %d",&hp,&wp,&hm,&wm))
{
getchar();
for(int i=0;i<hp;++i) gets(G1[i]);
for(int i=0;i<hm;++i) gets(G2[i]);
GetHash();
printf("%d ",solve());
}
return 0;
}
/*
*/
hash+KMP:
/*
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-08-09-13.27
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#define LL long long
#define ULL unsigned long long
using namespace std;
const int MAXN = 2010;
const ULL seed =1000000007;
ULL base[MAXN];
ULL hashval[MAXN][MAXN],Ti[MAXN][MAXN],hashTmp[MAXN],val[MAXN];
int Next[MAXN],h,w,n,m;
char s[MAXN][MAXN],t[MAXN][MAXN];
int KMP()
{
int ans=0;
for(int i=0,j=-1;i<m;++i)
{
while(j!=-1 && val[i]!=hashTmp[j+1]) j=Next[j];
if(val[i]==hashTmp[j+1])
{
if(++j==w-1)
{
j=Next[j];
ans++;
}
}
}
return ans;
}
void init(char s[MAXN][MAXN],ULL hashval[MAXN][MAXN],int h,int w)
{
for(int j=0;j<w;++j)
{
hashval[h][j]=0;
for(int i=h-1;i>=0;--i) hashval[i][j]=hashval[i+1][j]*seed+s[i][j];
}
}
int solve()
{
init(s,hashval,h,w);
init(t,Ti,n,m);
for(int i=0;i<w;++i)
{
hashTmp[i]=hashval[0][i];
}
// KMP Matching
Next[0]=-1;
// puts("- - - - - - - - - - - Program Run Here ! - - - - - - - - - - - - ");
for(int i=1,j=-1;i<w;++i)
{
while(j!=-1 && hashTmp[i]!=hashTmp[j+1]) j=Next[j];
Next[i]=hashTmp[i]==hashTmp[j+1]? ++j:j;
}
// for(int i=1;i<w;++i)
// {
// printf("%d",Next[i]);
// }
int ans=0;
for(int i=0;i<n-h+1;++i)
{
for(int j=0;j<m;++j)
val[j]=Ti[i][j]-Ti[i+h][j]*base[h];
ans+=KMP();
}
return ans;
}
void _()
{
base[0]=1;
for(int i=1;i<MAXN;++i) base[i]=base[i-1]*seed;
}
int main()
{
_();
while(~scanf("%d %d %d %d",&h,&w,&n,&m))
{
getchar();
for(int i=0;i<h;++i) gets(s[i]);
for(int i=0;i<n;++i) gets(t[i]);
printf("%d ",solve());
}
}
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-08-09-13.27
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#define LL long long
#define ULL unsigned long long
using namespace std;
const int MAXN = 2010;
const ULL seed =1000000007;
ULL base[MAXN];
ULL hashval[MAXN][MAXN],Ti[MAXN][MAXN],hashTmp[MAXN],val[MAXN];
int Next[MAXN],h,w,n,m;
char s[MAXN][MAXN],t[MAXN][MAXN];
int KMP()
{
int ans=0;
for(int i=0,j=-1;i<m;++i)
{
while(j!=-1 && val[i]!=hashTmp[j+1]) j=Next[j];
if(val[i]==hashTmp[j+1])
{
if(++j==w-1)
{
j=Next[j];
ans++;
}
}
}
return ans;
}
void init(char s[MAXN][MAXN],ULL hashval[MAXN][MAXN],int h,int w)
{
for(int j=0;j<w;++j)
{
hashval[h][j]=0;
for(int i=h-1;i>=0;--i) hashval[i][j]=hashval[i+1][j]*seed+s[i][j];
}
}
int solve()
{
init(s,hashval,h,w);
init(t,Ti,n,m);
for(int i=0;i<w;++i)
{
hashTmp[i]=hashval[0][i];
}
// KMP Matching
Next[0]=-1;
// puts("- - - - - - - - - - - Program Run Here ! - - - - - - - - - - - - ");
for(int i=1,j=-1;i<w;++i)
{
while(j!=-1 && hashTmp[i]!=hashTmp[j+1]) j=Next[j];
Next[i]=hashTmp[i]==hashTmp[j+1]? ++j:j;
}
// for(int i=1;i<w;++i)
// {
// printf("%d",Next[i]);
// }
int ans=0;
for(int i=0;i<n-h+1;++i)
{
for(int j=0;j<m;++j)
val[j]=Ti[i][j]-Ti[i+h][j]*base[h];
ans+=KMP();
}
return ans;
}
void _()
{
base[0]=1;
for(int i=1;i<MAXN;++i) base[i]=base[i-1]*seed;
}
int main()
{
_();
while(~scanf("%d %d %d %d",&h,&w,&n,&m))
{
getchar();
for(int i=0;i<h;++i) gets(s[i]);
for(int i=0;i<n;++i) gets(t[i]);
printf("%d ",solve());
}
}