poj 2019 Cornfields (二维rmq)
题意：给一个方阵，k个查询，每个查询求某个方阵的最大值和最小值之差。
思路：二维rmq.同时用到最大值和最小值的话可以把初始化写在一起。
/* ***********************************************
Author :111qqz
Created Time :2016年05月16日 星期一 18时31分23秒
File Name :code/poj/2019.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=251;
7int a[N][N];
8int dp[N][N][8][8];
9int dp2[N][N][8][8];
10int n,b,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][0] = dp2[i][j][0][0] = a[i][j];
1 for ( int i = 0 ; (1<<i)<= n ; i++)
2 for ( int j = 0 ; (1<<j) <= n ; 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 <= n ; q++)
6 if (i==0)
7 {
8 dp[p][q][i][j] = max(dp[p][q][i][j-1],dp[p][q+(1<<(j-1))][i][j-1]);
9 dp2[p][q][i][j] = min(dp2[p][q][i][j-1],dp2[p][q+(1<<(j-1))][i][j-1]);
10 }
11 else
12 {
13 dp[p][q][i][j] = max(dp[p][q][i-1][j],dp[p+(1<<(i-1))][q][i-1][j]);
14 dp2[p][q][i][j] = min(dp2[p][q][i-1][j],dp2[p+(1<<(i-1))][q][i-1][j]);
15 }
16}
1int _rmq(int x1,int y1,int x2,int y2)
2{
3 int k1 = 0 ;
4 int k2 = 0 ;
5 while (1<<(k1+1)<=x2-x1+1) k1++;
6 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];
int mx = max(max(tmp1,tmp2),max(tmp3,tmp4));
1 tmp1 = dp2[x1][y1][k1][k2];
2 tmp2 = dp2[x2-(1<<k1)+1][y1][k1][k2];
3 tmp3 = dp2[x1][y2-(1<<k2)+1][k1][k2];
4 tmp4 = dp2[x2-(1<<k1)+1][y2-(1<<k2)+1][k1][k2];
int mn = min(min(tmp1,tmp2),min(tmp3,tmp4));
// cout<<"mx:"<<mx<<" mn:"<<mn<<endl;
return mx - mn;
}
1int main()
2{
3 #ifndef ONLINE_JUDGE
4 freopen("code/in.txt","r",stdin);
5 #endif
6 scanf("%d %d %d",&n,&b,&q);
7 for ( int i = 1 ; i <= n ; i++)
8 for ( int j = 1 ; j <= n ; j++) scanf("%d",&a[i][j]);
9 init_rmq();
1 while (q--)
2 {
3 int x1,y1;
4 scanf("%d %d",&x1,&y1);
5 printf("%d\n",_rmq(x1,y1,x1+b-1,y1+b-1));
6 }
1 #ifndef ONLINE_JUDGE
2 fclose(stdin);
3 #endif
4 return 0;
5}