HNU13383:The Big Painting

最新推荐文章于 2020-05-13 17:02:13 发布

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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

Output

A single integer representing the number of possible locations where his painting might be.

Sample Input

4 4 10 10
oxxo
xoox
xoox
oxxo
xxxxxxoxxo
oxxoooxoox
xooxxxxoox
xooxxxoxxo
oxxoxxxxxx
ooooxxxxxx
xxxoxxoxxo
oooxooxoox
oooxooxoox
xxxoxxoxxo

Sample Output

Problem Source

HNU Contest

题意：

给出两个图，问大图中包含几个小图

思路：

用数字保存是一种方法，本以为超类型肯定不行的，没有想到居然也能AC，不知道是数据水还是数据保存方面的一些玄机？

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;

#define ls 2*i
#define rs 2*i+1
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define ULL unsigned long long
#define N 2005
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define rank rank1
const int mod = 1000000007;

int h1,h2,w1,w2,ans;
char s1[N][N],s2[N][N];
const ULL t1 = 7;
const ULL t2 = 11;
ULL sum,a[N][N],b[N][N];

void solve()//把大图用数字表示的同时找出与小图相等的位置
{
    int i,j,k;
    ULL x=1;
    for(i = 0; i<w1; i++) x*=t1;
    for(i = 0; i<h2; i++)
    {
        ULL s = 0;
        for(j = 0; j<w1; j++) s = s*t1+s2[i][j];
        a[i][w1-1] = s;
        for(j = w1; j<w2; j++)
            a[i][j] = a[i][j-1]*t1-s2[i][j-w1]*x+s2[i][j];
    }
    x = 1;
    for(i = 0; i<h1; i++) x*=t2;
    for(i = w1-1; i<w2; i++)
    {
        ULL s = 0;
        for(j = 0; j<h1; j++) s = s*t2+a[j][i];
        b[h1-1][i] = s;
        if(s==sum) ans++;
        for(j = h1; j<h2; j++)
        {
            b[j][i] = b[j-1][i]*t2-a[j-h1][i]*x+a[j][i];
            if(b[j][i]==sum) ans++;
        }
    }
}

int main()
{
    int i,j,k;
    while(~scanf("%d%d%d%d",&h1,&w1,&h2,&w2))
    {
        for(i = 0; i<h1; i++) scanf("%s",s1[i]);
        for(i = 0; i<h2; i++) scanf("%s",s2[i]);
        sum = 0;
        for(i = 0; i<h1; i++)//把小图用数字表示
        {
            ULL s = 0;
            for(j = 0; j<w1; j++)
            {
                s = s*t1+s1[i][j];
            }
            sum = sum*t2+s;
        }
        ans = 0;
        solve();
        printf("%d\n",ans);
    }

    return 0;
}