ZOJ 题目2859 Matrix Searching（二维RMQ）

Posted 2020-09-19 zhchoutai

Matrix Searching

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Time Limit: 10 Seconds      Memory Limit: 32768 KB

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Given an n*n matrix A, whose entries Ai,j are integer numbers ( 1 <= i <= n, 1 <= j <= n ). An operation FIND the minimun number in a given ssub-matrix.

Input

The first line of the input contains a single integer T , the number of test cases.

For each test case, the first line contains one integer n (1 <= n <= 300), which is the sizes of the matrix, respectively. The next n lines with n integers each gives the elements of the matrix.

The next line contains a single integer N (1 <= N <= 1,000,000), the number of queries. The next N lines give one query on each line, with four integers r1, c1, r2, c2 (1 <= r1 <= r2 <= n, 1 <= c1 <= c2 <= n), which are the indices of the upper-left corner and lower-right corner of the sub-matrix in question.

Output

For each test case, print N lines with one number on each line, the required minimum integer in the sub-matrix.

Sample Input

1
2
2 -1
2 3
2
1 1 2 2
1 1 2 1

Sample Output

-1
2

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Author: PENG, Peng

Source: ZOJ Monthly, June 2007

帮学妹找了一晚上的bug。，，各种调试，。各种报错，。感觉自己水爆了，，赶紧水道高级点的水题压压惊~~

ac代码

#include<stdio.h>
#include<string.h>
#include<math.h>
#define max(a,b) (a>b?a:b)
#define min(a,b) (a>b?b:a)
int map[301][301];
int minv[301][301][9][9];
int mm[306];
int n,m;
void initrmq()
{
	int i,j;
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=n;j++)
		{
			minv[i][j][0][0]=map[i][j];
		}
	}
	int ii,jj;
	for(ii=0;ii<=mm[n];ii++)
		for(jj=0;jj<=mm[n];jj++)
		{
			if(ii+jj)
			{
				for(i=1;i+(1<<ii)-1<=n;i++)
					for(j=1;j+(1<<jj)-1<=n;j++)
					{
						if(ii)
							minv[i][j][ii][jj]=min(minv[i][j][ii-1][jj],minv[i+(1<<(ii-1))][j][ii-1][jj]);
						else
							minv[i][j][ii][jj]=min(minv[i][j][ii][jj-1],minv[i][j+(1<<(jj-1))][ii][jj-1]);
					}
			}
		}
}
int q_min(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(minv[x1][y1][k1][k2],minv[x1][y2][k1][k2]),min(minv[x2][y1][k1][k2],minv[x2][y2][k1][k2]));
}
void init()
{
	mm[0]=-1;
	int i;
  for(i = 1;i <= 305;i++)  
        mm[i] = ((i&(i-1))==0)?mm[i-1]+1:mm[i-1];  

}
int main()
{
	int t;
	init();
	scanf("%d",&t);
	while(t--)
	{
		//int n,m;
		scanf("%d",&n);
		int i,j;
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=n;j++)
				scanf("%d",&map[i][j]);
		}
		initrmq();
		scanf("%d",&m);
		while(m--)
		{
			int r1,c1,r2,c2;
			scanf("%d%d%d%d",&r1,&c1,&r2,&c2);
			printf("%d\n",q_min(r1,c1,r2,c2));
		}
	}
}