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?

 1/* ***********************************************
 2Author :111qqz
 3Created Time :2016年05月16日 星期一 19时24分26秒
 4File Name :code/loj/1081.cpp
 5************************************************ */
 6
 7#include <cstdio>
 8#include <cstring>
 9#include <iostream>
10#include <algorithm>
11#include <vector>
12#include <queue>
13#include <set>
14#include <map>
15#include <string>
16#include <cmath>
17#include <cstdlib>
18#include <ctime>
19#define fst first
20#define sec second
21#define lson l,m,rt<<1
22#define rson m+1,r,rt<<1|1
23#define ms(a,x) memset(a,x,sizeof(a))
24typedef long long LL;
25#define pi pair < int ,int >
26#define MP make_pair
27
28using namespace std;
29const double eps = 1E-8;
30const int dx4[4]={1,0,0,-1};
31const int dy4[4]={0,-1,1,0};
32const int inf = 0x3f3f3f3f;
33const int N=501;
34int a[N][N];
35int dp[N][N][9];
36int n,q;
37
38void init_rmq()
39{
40    for ( int i = 1 ; i <= n ; i++)
41	for ( int j = 1 ; j <= n ; j++)
42	    dp[i][j][0] = a[i][j];
43
44    for ( int i = 1 ; (1<<i)<= n ; i++)
45	for ( int p = 1 ; p + (1<<i)-1 <= n ; p++)
46	    for ( int  q = 1 ; q + (1<<i)-1 <= n ; q++)
47		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]));
48
49
50
51}
52
53int rmq_max(int x1,int y1,int x2,int y2)
54{
55    int k = 0;
56    while (1<<(k+1)<= x2-x1+1) k++;
57
58    int tmp1 = dp[x1][y1][k];
59    int tmp2 = dp[x2-(1<<k)+1][y1][k];
60    int tmp3 = dp[x1][y2-(1<<k)+1][k];
61    int tmp4 = dp[x2-(1<<k)+1][y2-(1<<k)+1][k];
62
63    return max(max(tmp1,tmp2),max(tmp3,tmp4));
64}
65int main()
66{
67	#ifndef  ONLINE_JUDGE 
68	freopen("code/in.txt","r",stdin);
69  #endif
70
71	int T;
72	scanf("%d",&T);
73	int cas = 0 ;
74	while (T--)
75	{
76	    printf("Case %d:\n",++cas);
77	    scanf("%d %d",&n,&q);
78	    ms(a,0);
79	    for ( int i = 1 ; i <= n ; i++)
80		for ( int j = 1 ; j <= n ; j++) scanf("%d",&a[i][j]);
81
82	    init_rmq();
83
84	    while (q--)
85	    {
86		int x,y,s;
87		scanf("%d %d %d",&x,&y,&s);
88
89		printf("%d\n",rmq_max(x,y,x+s-1,y+s-1));
90	    }
91	}
92
93  #ifndef ONLINE_JUDGE  
94  fclose(stdin);
95  #endif
96    return 0;
97}