【POJ 2019]二维RMQ

1.​​题目链接​​。题目大意，给定一个矩阵，k组询问，每组询问(x,y)输出以(x,y)为左上角，宽度为B的子矩阵里面最大值和最小值之差。

2.二维RMQ的裸题了，dp[i][j][a][b]代表的是当前点在(i,j)到(i+pow(2,a),j+pow(2,b))的最值。询问也是一样。模板题。

#include<iostream>
#include <string.h>
#include <iostream>
#include <algorithm>
#include<stdio.h>
#include<cstdio>
using namespace std;

int val[255][255];
int mm[255];
int dpmin[255][255][8][8];//最小值
int dpmax[255][255][8][8];//最大值

void initRMQ(int n, int m)
{
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
dpmin[i][j][0][0] = dpmax[i][j][0][0] = val[i][j];
for (int ii = 0; ii <= mm[n]; ii++)
for (int jj = 0; jj <= mm[m]; jj++)
if (ii + jj)
for (int i = 1; i + (1 << ii) - 1 <= n; i++)
for (int j = 1; j + (1 << jj) - 1 <= m; j++)
{
if (ii)
{
dpmin[i][j][ii][jj] = min(dpmin[i][j][ii - 1][jj], dpmin[i + (1 << (ii - 1))][j][ii - 1][jj]);
dpmax[i][j][ii][jj] = max(dpmax[i][j][ii - 1][jj], dpmax[i + (1 << (ii - 1))][j][ii - 1][jj]);
}
else
{
dpmin[i][j][ii][jj] = min(dpmin[i][j][ii][jj - 1], dpmin[i][j + (1 << (jj - 1))][ii][jj - 1]);
dpmax[i][j][ii][jj] = max(dpmax[i][j][ii][jj - 1], dpmax[i][j + (1 << (jj - 1))][ii][jj - 1]);
}
}
}
//查询矩形的最大值
int rmq1(int x1, int y1, int x2, int y2)
{
int k1 = mm[x2 - x1 + 1];
int k2 = mm[y2 - y1 + 1];
x2 = x2 - (1 << k1) + 1;
y2 = y2 - (1 << k2) + 1;
return max(max(dpmax[x1][y1][k1][k2], dpmax[x1][y2][k1][k2]), max(dpmax[x2][y1][k1][k2], dpmax[x2][y2][k1][k2]));
}



int rmq2(int x1, int y1, int x2, int y2)
{
int k1 = mm[x2 - x1 + 1];
int k2 = mm[y2 - y1 + 1];
x2 = x2 - (1 << k1) + 1;
y2 = y2 - (1 << k2) + 1;
return min(min(dpmin[x1][y1][k1][k2], dpmin[x1][y2][k1][k2]), min(dpmin[x2][y1][k1][k2], dpmin[x2][y2][k1][k2]));
}


int main()
{
mm[0] = -1;
for (int i = 1; i <= 500; i++)
mm[i] = ((i&(i - 1)) == 0) ? mm[i - 1] + 1 : mm[i - 1];
int N, B, K;
while (scanf("%d%d%d", &N, &B, &K) == 3)
{
for (int i = 1; i <= N; i++)
for (int j = 1; j <= N; j++)
scanf("%d", &val[i][j]);
initRMQ(N, N);
int x, y;
while (K--)
{
scanf("%d%d", &x, &y);
printf("%d\n", rmq1(x, y, x + B - 1, y + B - 1) - rmq2(x, y, x + B - 1, y + B - 1));
}
}
return 0;
}

 

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