HOJ 13383 The Big Painting 二维字符串hash

The Big PaintingTime Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KBTotal submit users: 37, Accepted users: 17Problem 13383 : No special judgementProblem 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?

给你两个字符矩阵，求一个字符矩阵在另一个矩阵里出现的次数

直接搜了一个二维字符串hash的板子，交上去了

#include<cstdio>#include<iostream>using namespace std;#define maxn 2005#define ULL unsigned long longULL maxSize = 4000007;ULL seedx = 233;ULL seedy = 23333;struct HashString {ULL xp[maxn];ULL yp[maxn];ULL H[maxn][maxn];char s[maxn][maxn];int x,y;void init(int n, int m) {xp[0] = yp[0] = 1;x = n,y = m;for(int i = 1; i <= x; i++) xp[i] = xp[i - 1]*seedx;for(int i = 1; i <= y; i++) yp[i] = yp[i - 1]*seedy;return;}void initHash() {for(int i = 0; i < x; i++) {H[i][0] = s[i][0] - 'A' + 1;for(int j = 1; j < y; j++)H[i][j] = H[i][j - 1]*seedy + s[i][j] - 'A' + 1;}for(int i = 1; i < x; i++)//从1开始, 第0行是不变的for(int j = 0; j < y; j++)H[i][j] += H[i - 1][j]*seedx;return;}ULL askHash(int x1, int y1, int x2, int y2) { //询问以(x1, y1)为左上角, (x2, y2)为右上角的字符矩阵的hash值ULL ret = H[x2][y2];if(x1 > 0) ret -= H[x1 - 1][y2]*xp[x2 - x1 + 1];if(y1 > 0) ret -= H[x2][y1 - 1]*yp[y2 - y1 + 1];if(x1 > 0 && y1 > 0) ret += H[x1 - 1][y1 - 1]*xp[x2 - x1 + 1]*yp[y2 - y1 + 1];return ret;}void read(){    for(int i  =0;i<x;i++)            scanf("%s",s[i]);}}s1,s2;int main() {    #ifdef ACBang        freopen("1.in","r",stdin);    #endif // ACBang    int x,y,n,m;    while(scanf("%d%d%d%d",&x,&y,&n,&m)!=EOF)    {        s1.init(x,y);        s1.read();        s2.init(n,m);        s2.read();        s1.initHash();        s2.initHash();        ULL tt = s1.askHash(0,0,x-1,y-1);        int ans = 0;        for(int i = 0;i<n-x+1;i++)            {                for(int j = 0;j<m-y+1;j++)                {                    if(tt == s2.askHash(i,j,i+x-1,j+y-1))                      ans ++;                }            }        printf("%d\n",ans);    }return 0;}