UvaLive6893_The_Big_Painting

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Catalog

目录

• Catalog
  • Problem：传送门
• Solution：
  • AC_Code：
  • Problem Description：

Problem：传送门

 Portal

 原题目描述在最下面。

 给你两个二维矩阵，问第一个矩阵在第二个矩阵中的出现次数。

Solution：

二维hash:

 直接二维矩阵hash，枚举求值即可。注意横纵base值不要取相同。枚举的时候注意一些小细节。

hash+Kmp:

 一维hash，把一个矩阵hash成一维序列。然后另一位维Kmp判断出现次数。

AC_Code：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef unsigned long long uLL;

const uLL base1 = 1572872831;
const uLL base2 = 1971536491;
const int MXN = 2005;
int n1, n2, m1, m2;
char ar[MXN][MXN], br[MXN][MXN];
uLL cr[MXN][MXN];
int solve(int n1,int m1,int n2,int m2) {
    int cnt = 0;
    uLL ans1 = 0, tmp, pw1 = 1, pw2 = 1;
    for(int i = 1; i <= m1; ++i) pw1 = pw1 * base1;
    for(int i = 1; i <= n1; ++i) {
        tmp = 0;
        pw2 = pw2 * base2;
        for(int j = 1; j <= m1; ++j) {
            tmp = tmp * base1 + ar[i][j];
        }
        ans1 = ans1 * base2 + tmp;
    }//ans1是第一个矩阵的hash值
    for(int i = 1; i <= n2; ++i) {
        for(int j = 1; j <= m1; ++j) {
            cr[i][j] = cr[i][j-1] * base1 + br[i][j];
        }
        for(int j = m1+1; j <= m2; ++j) {//预处理第i行第j个字母前m1的字母的一维hash值
            cr[i][j] = cr[i][j-1] * base1 + br[i][j] - br[i][j-m1]*pw1;
        }
    }
    for(int j = m1; j <= m2; ++j) {//枚举列
        tmp = 0;
        for(int i = 1; i <= n1; ++i) tmp = tmp * base2 + cr[i][j];
        if(tmp == ans1) cnt++;
        for(int i = n1 + 1; i <= n2; ++i) {//维持长度为n1
            tmp = tmp * base2 + cr[i][j] - cr[i-n1][j]*pw2;
            if(tmp == ans1) cnt++;
        }
    }
    return cnt;
}
int main(){
    while(~scanf("%d%d%d%d", &n1, &m1, &n2, &m2)){
        for(int i = 1; i <= n1; ++i) scanf("%s", ar[i]+1);
        for(int i = 1; i <= n2; ++i) scanf("%s", br[i]+1);
        printf("%d\n", solve(n1,m1,n2,m2));
    }
    return 0;
}
/*
4 4 10 10
oxxo
xoox
xoox
oxxo
xxxxxxoxxo
oxxoooxoox
xooxxxxoox
xooxxxoxxo
oxxoxxxxxx
ooooxxxxxx
xxxoxxoxxo
oooxooxoox
oooxooxoox
xxxoxxoxxo
*/

Problem Description：

Samuel W. E. R. Craft is an artist with a growing reputation.

Unfortunately, the paintings he sells do not provide

him enough money for his daily expenses plus the new supplies

he needs. He had a brilliant idea yesterday when he

ran out of blank canvas: ”Why don’t I create a gigantic

new painting, made of all the unsellable paintings I have,

stitched together?”. After a full day of work, his masterpiece

was complete.

That’s when he received an unexpected phone call: a

client saw a photograph of one of his paintings and is willing

to buy it now! He had forgotten to tell the art gallery to

remove his old works from the catalog! He would usually

welcome a call like this, but how is he going to find his old

work in the huge figure in front of him?

Given a black-and-white representation of his original

painting and a black-and-white representation of his masterpiece, can you help S.W.E.R.C. identify in

how many locations his painting might be?

Input

The input file contains several test cases, each of them as described below.

The first line consists of 4 space-separated integers: hp wp hm wm, the height and width of the

painting he needs to find, and the height and width of his masterpiece, respectively.

The next hp lines have wp lower-case characters representing his painting. After that, the next hm

lines have wm lower-case characters representing his masterpiece. Each character will be either ‘x’ or

‘o’.

Constraints:

1 ≤ hp, wp ≤ 2 000

1 ≤ hm, wm ≤ 2 000

hp ≤ hm

wp ≤ wm

Output

For each test case, output a single integer representing the number of possible locations where his

painting might be, on a line by itself.

Sample Output Explanation

The painting could be in four locations as shown in the following picture. Two of the locations overlap.

Sample Input

4 4 10 10

oxxo

xoox

xoox

oxxo

xxxxxxoxxo

oxxoooxoox

xooxxxxoox

xooxxxoxxo

oxxoxxxxxx

ooooxxxxxx

xxxoxxoxxo

oooxooxoox

oooxooxoox

xxxoxxoxxo

Sample Output

4