lightoj 1081 Square Queries (二维rmq,降维)

lightoj 1081 题目链接

题意:和上一道一样,但是由于size变成了500,如果按照之前的做法会tle + mle...

很容易发现,由于是方阵,长宽是相等的,所以有一维是可以省略的。

也就是所谓的降维?

/* ***********************************************
Author :111qqz
Created Time :2016年05月16日 星期一 19时24分26秒
File Name :code/loj/1081.cpp
************************************************ */
 1#include <cstdio>
 2#include <cstring>
 3#include <iostream>
 4#include <algorithm>
 5#include <vector>
 6#include <queue>
 7#include <set>
 8#include <map>
 9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6const int N=501;
7int a[N][N];
8int dp[N][N][9];
9int n,q;
1void init_rmq()
2{
3    for ( int i = 1 ; i <= n ; i++)
4	for ( int j = 1 ; j <= n ; j++)
5	    dp[i][j][0] = a[i][j];
1    for ( int i = 1 ; (1<<i)<= n ; i++)
2	for ( int p = 1 ; p + (1<<i)-1 <= n ; p++)
3	    for ( int  q = 1 ; q + (1<<i)-1 <= n ; q++)
4		dp[p][q][i] = max(max(dp[p][q][i-1],dp[p+(1<<(i-1))][q][i-1]),max(dp[p][q+(1<<(i-1))][i-1],dp[p+(1<<(i-1))][q+(1<<(i-1))][i-1]));
}
1int rmq_max(int x1,int y1,int x2,int y2)
2{
3    int k = 0;
4    while (1<<(k+1)<= x2-x1+1) k++;
1    int tmp1 = dp[x1][y1][k];
2    int tmp2 = dp[x2-(1<<k)+1][y1][k];
3    int tmp3 = dp[x1][y2-(1<<k)+1][k];
4    int tmp4 = dp[x2-(1<<k)+1][y2-(1<<k)+1][k];
1    return max(max(tmp1,tmp2),max(tmp3,tmp4));
2}
3int main()
4{
5	#ifndef  ONLINE_JUDGE 
6	freopen("code/in.txt","r",stdin);
7  #endif
 1	int T;
 2	scanf("%d",&T);
 3	int cas = 0 ;
 4	while (T--)
 5	{
 6	    printf("Case %d:\n",++cas);
 7	    scanf("%d %d",&n,&q);
 8	    ms(a,0);
 9	    for ( int i = 1 ; i <= n ; i++)
10		for ( int j = 1 ; j <= n ; j++) scanf("%d",&a[i][j]);
	    init_rmq();
1	    while (q--)
2	    {
3		int x,y,s;
4		scanf("%d %d %d",&x,&y,&s);
1		printf("%d\n",rmq_max(x,y,x+s-1,y+s-1));
2	    }
3	}
1  #ifndef ONLINE_JUDGE  
2  fclose(stdin);
3  #endif
4    return 0;
5}