hdu 2888 check corners (二维rmq模板题)

2016-05-16 · 1 min read · rmq

hdu2888题目链接

题意：问某个矩阵内的最大值，并且问最大值是否是在四个角中出现。 思路：二维rmq.需要注意数组稍微开大1就会MLE,因为是四维数组，一维大一点，整个就会大很多==。

/* ***********************************************
Author :111qqz
Created Time :2016年05月16日 星期一 16时51分00秒
File Name :code/hdu/2888.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=301;
7int n,m,q;
8int a[N][N];
9int dp[N][N][9][9];

1void init_rmq()
2{
3    for ( int i = 1 ; i <= n ; i++)
4	for ( int j = 1; j <= m ; j++) dp[i][j][0][0] = a[i][j];

1    for ( int i = 0 ; (1<<i)<= n ; i++)
2	for ( int j = 0 ; (1<<j)<= m  ;j++)
3	    if (i==0&&j==0) continue;
4	    else for ( int p = 1 ; p + (1<<i)-1 <= n ; p++)
5		      for ( int q = 1 ; q + (1<<j)-1 <= m ;q ++)
6			  if (i==0)
7			      dp[p][q][i][j] = max(dp[p][q][i][j-1],dp[p][q+(1<<(j-1))][i][j-1]);
8			  else dp[p][q][i][j] = max(dp[p][q][i-1][j],dp[p+(1<<(i-1))][q][i-1][j]);

}

1int rmq_max(int x1,int y1,int x2,int y2)
2{
3    int k1 = 0 ;
4    int k2 = 0 ;

    while (1<<(k1+1)<=x2-x1+1) k1++;
    while (1<<(k2+1)<=y2-y1+1) k2++;

1    int tmp1 = dp[x1][y1][k1][k2];
2    int tmp2 = dp[x2-(1<<k1)+1][y1][k1][k2];
3    int tmp3 = dp[x1][y2-(1<<k2)+1][k1][k2];
4    int tmp4 = dp[x2-(1<<k1)+1][y2-(1<<k2)+1][k1][k2];

    return max(max(tmp1,tmp2),max(tmp3,tmp4));
}

1int main()
2{
3	#ifndef  ONLINE_JUDGE 
4	freopen("code/in.txt","r",stdin);
5  #endif

1	while (scanf("%d %d",&n,&m)!=EOF)
2	{
3	    for ( int i = 1 ; i <= n ; i++)
4		for ( int j = 1 ; j <= m ; j++) scanf("%d",&a[i][j]);
5	    init_rmq();

 1	    scanf("%d",&q);
 2	    while (q--)
 3	    {
 4		int r1,c1,r2,c2;
 5		scanf("%d %d %d %d",&r1,&c1,&r2,&c2);
 6		int ans = rmq_max(r1,c1,r2,c2);
 7		printf("%d ",ans);
 8		if (a[r1][c1]==ans||a[r1][c2]==ans||a[r2][c1]==ans||a[r2][c2]==ans)
 9		{
10		    puts("yes");
11		}
12		else
13		{
14		    puts("no");
15		}
16	    }

	}

1  #ifndef ONLINE_JUDGE  
2  fclose(stdin);
3  #endif
4    return 0;
5}