POJ 2019 Cornfields

 

Cornfields

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 3096		Accepted: 1459

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

Source

USACO 2003 March Green

 

/* DP,其实对每一行DP即可,当前行当前列的状态只与同一行的前一列的状态有关,因此采用二维滚动 数组temp来计算以节约空间,temp的第一维用来滚动表示当前列和前一列的状态 temp[0 or 1][cp][0] temp[0 or 1][cp][1],分别表示当前行中以当前列结尾的长度为cp的数列的最 小值和最大值. 那么: temp[i][cp][0] = min(temp[(i + 1) % 2][cp - 1][0], val); val 是当前行当前列的输入值; temp[i][cp][1] = max(temp[(i + 1) % 2][cp - 1][1], val); 每处理完一个点[i,j],将temp[i][cp][0] 和temp[i][cp][1]记录到tarray[i][j][0]和tarray[i][j][1]中 最后处理查询请求的时候,对于请求[x][y],先确定这个矩形范围的最右下角的坐标为[x + b - 1][y + b - 1] 然后分别扫描从这个矩形的b行,从其每行最后一个元素的tarray[][][0]和tarray[][][1]中找出最大值和最小值 即可,即最后的结果是 minval = min(minval, tarray[x + i][y + b - 1][0]), 0 <= i < b maxval = max(maxval, tarray[x + i][y + b - 1][1]), 0 <= i < b 算法的时间复杂度是O(N ^ 2 * B + KB),是否有更快的算法留着以后研究 */ #include <iostream> #define MAX_N 250 #define maxv(a, b) ((a) >= (b) ? (a) : (b)) #define minv(a, b) ((a) <= (b) ? (a) : (b)) using namespace std; int tarray[MAX_N + 1][MAX_N + 1][2]; int temp[2][MAX_N + 1][2]; int n, b, k; int main() { int i, j; scanf("%d%d%d", &n, &b, &k); //初始化边界条件 temp[0][0][0] = temp[1][0][0] = INT_MAX; temp[0][0][1] = temp[1][0][1] = INT_MIN; //扫描每一行 for(i = 1; i <= n; i++) { //扫描每一列 for(j = 1; j <= n; j++) { int val; scanf("%d", &val); //第一列单独处理一下 if(j == 1) { temp[0][1][0] = tarray[i][j][0] = val; temp[0][1][1] = tarray[i][j][1] = val; continue; } //kp是以当前列结尾可以处理的数列的最大长度与b的最小值 int kp = minv(j, b); //计算以当前列元素结尾的长度为cp的数列的最大值和最小值 for(int cp = kp; cp >= 1; cp--) { temp[(j + 1) % 2][cp][0] = minv(temp[j % 2][cp - 1][0], val); temp[(j + 1) % 2][cp][1] = maxv(temp[j % 2][cp - 1][1], val); } //当可以构成大于等于b长度的数列需要记忆一下 if(j >= b) { tarray[i][j][0] = temp[(j + 1) % 2][b][0]; tarray[i][j][1] = temp[(j + 1) % 2][b][1]; } } } //处理查询请求 int row, col; for(i = 1; i <= k; i++) { int maxval = INT_MIN, minval = INT_MAX; scanf("%d%d", &row, &col); row = row + b - 1; col = col + b - 1; for(j = row; j >= row - b + 1; j--) { maxval = maxv(maxval, tarray[j][col][1]); minval = minv(minval, tarray[j][col][0]); } printf("%d/n", maxval - minval); } return 0; }