POJ2019Cornfields 题解

Cornfields
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 6364 Accepted: 3131
Description

FJ has decided to grow his own corn hybrid in order to help the cows make the best possible milk. To that end, he’s looking to build the cornfield on the flattest piece of land he can find.

FJ has, at great expense, surveyed his square farm of N x N hectares (1 <= N <= 250). Each hectare has an integer elevation (0 <= elevation <= 250) associated with it.

FJ will present your program with the elevations and a set of K (1 <= K <= 100,000) queries of the form “in this B x B submatrix, what is the maximum and minimum elevation?”. The integer B (1 <= B <= N) is the size of one edge of the square cornfield and is a constant for every inquiry. Help FJ find the best place to put his cornfield.
Input

• Line 1: Three space-separated integers: N, B, and K.

• Lines 2..N+1: Each line contains N space-separated integers. Line 2 represents row 1; line 3 represents row 2, etc. The first integer on each line represents column 1; the second integer represents column 2; etc.

• Lines N+2..N+K+1: Each line contains two space-separated integers representing a query. The first integer is the top row of the query; the second integer is the left column of the query. The integers are in the range 1..N-B+1.
  Output

• Lines 1..K: A single integer per line representing the difference between the max and the min in each query.
  Sample Input

5 3 1
5 1 2 6 3
1 3 5 2 7
7 2 4 6 1
9 9 8 6 5
0 6 9 3 9
1 2
Sample Output

5
中文大意：
题面
给出一个N*N （N<=250）的方阵，以及K（<=100000）个询问。每次询问如下：以（Xi,Yi）为左上角，边长为B的子方阵中，最大值和最小值的差是多少？

注意对于所有的询问，B都是一个定值。

输入
第一行N，B（<=N），K。含义如上。

接下来N行N列的一个矩阵，每个数<=250。

接下来K行表示询问，每行两个数Xi, Yi 表示询问的方阵的左上角。

输出
一行一个正整数，含义如上。

样例
输入
5 3 1
5 1 2 6 3
1 3 5 2 7
7 2 4 6 1
9 9 8 6 5
0 6 9 3 9
1 2
输出
5
思路：
只有250*250，枚举+记忆化即可过

#include<stdio.h>int n,b,k,i,j,x,y,max,min,a[300][300],f[300][300];int main(){    scanf("%d%d%d",&n,&b,&k);    for(i=1;i<=n;i++)        for(j=1;j<=n;j++)            scanf("%d",&a[i][j]);    while(k--){        scanf("%d%d",&x,&y);        if(!f[x][y]){            max=0;            min=10000000;            for(i=1;i<=b;i++)                for(j=1;j<=b;j++){                    if(a[i+x-1][j+y-1]<min)min=a[i+x-1][j+y-1];                    if(a[i+x-1][j+y-1]>max)max=a[i+x-1][j+y-1];                }            f[x][y]=max-min;//取最大最小值，用f数组存起来        }        printf("%d\n",f[x][y]);//输出处理后的f数组    }    return 0;}