POJ 3264 Balanced Lineup RMQ问题的ST解法

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原文链接：http://www.cnblogs.com/372465774y/archive/2012/09/15/2687032.html

本文介绍了一种解决Range Minimum Query (RMQ)问题的有效方法——Sparse Table算法，并通过一个具体的例子进行了说明。该算法利用预处理的方式，在O(log N)的时间复杂度内解决了最大值和最小值查询的问题。

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 24349		Accepted: 11348
Case Time Limit: 2000MS

Description

For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.

Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.

Input

Line 1: Two space-separated integers, N and Q.
Lines 2.. N+1: Line i+1 contains a single integer that is the height of cow i
Lines N+2.. N+ Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.

Output

Lines 1.. Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.

Sample Input

Sample Output

6
3
0

Source

USACO 2007 January Silver

RMQ问题的ST解法、第一次用，1Y，呵呵

开学后课好多，课程任务好重、都没什么时间做题目、学算法、烦呀


#include <iostream>
#include <stdio.h>
#define N 50010
#include <cmath>
using namespace std;
int f_min[N][18];
int f_max[N][18];

int main()
{
   // cout<<(1<<16);
   int n,q;
   int i,j,k,t;
   int h,l;
  // printf("%d ",k);
   while(scanf("%d %d",&n,&q)!=EOF)
   {
       for(i=1;i<=n;i++)
       {
           scanf("%d",&f_min[i][0]);
           f_max[i][0]=f_min[i][0];
       }
       k=(int)(log(n*1.0)/log(2*1.0));
       for(i=1;i<=k;i++)
          {
              t=(1<<i);
              for(j=1;j<=n-t+1;j++)
              {
                  f_max[j][i]=max(f_max[j][i-1],f_max[j+t/2][i-1]);
                  f_min[j][i]=min(f_min[j][i-1],f_min[j+t/2][i-1]);

              }
          }
       while(q--)
       {
           scanf("%d %d",&i,&j);
           k=(int)(log((j-i+1)*1.0)/log(2*1.0));
           t=(1<<k);
           h=max(f_max[i][k],f_max[j-t+1][k]);
           l=min(f_min[i][k],f_min[j-t+1][k]);
           printf("%d\n",h-l);
       }
   }
    return 0;
}

转载于:https://www.cnblogs.com/372465774y/archive/2012/09/15/2687032.html