POJ 3264-Balanced Lineup（RMQ-ST算法）

ゝ一纸荒年。 2022-06-18 10:54 192阅读 0赞

Balanced Lineup

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;5000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;51826</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;24285</td> 
  </tr> 
  <tr> 
   <td colspan="3" align="center" style="color:rgb(14,173,0)"><strong>Case Time Limit:</strong>&nbsp;2000MS</td> 
  </tr> 
 </tbody> 
</table>

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

6 3
    1
    7
    3
    4
    2
    5
    1 5
    4 6
    2 2

Sample Output

6
    3
    0

Source

[USACO 2007 January Silver][]

求N个数中给定区间L-R中最大值与最小值的差。

以前这题作为入门题用线段树搞了搞，突然翻出来用RMQ水了一发。

坑爹的是cin、cout会超时…

#include<iostream>
    #include<cstdio>
    #include<iomanip>
    #include<cmath>
    #include<cstdlib>
    #include<cstring>
    #include<map>
    #include<algorithm>
    #include<vector>
    #include<queue>
    using namespace std;
    #define INF 0x3f3f3f3f
    #define MAXN 50010
    int a[MAXN],dpmin[MAXN][20],dpmax[MAXN][20];
    int n,q;
    void rmq()
    {
        int temp=(int)(log((double)n)/log(2.0));
        for(int i=0; i<n; i++)
            dpmin[i][0]=a[i],dpmax[i][0]=a[i];
        for(int j=1; j<=temp; j++)
            for(int i=0; i<=n-(1<<j); i++)
            {
                dpmin[i][j]=min(dpmin[i][j-1],dpmin[i+(1<<(j-1))][j-1]);
                dpmax[i][j]=max(dpmax[i][j-1],dpmax[i+(1<<(j-1))][j-1]);
            }
    }
    int Minimum(int s,int e)
    {
        int k=(int)(log((double)e-s+1)/log(2.0));
        return min(dpmin[s][k],dpmin[e-(1<<k)+1][k]);
    }
    int Maxmum(int s,int e)
    {
        int k=(int)(log((double)e-s+1)/log(2.0));
        return max(dpmax[s][k],dpmax[e-(1<<k)+1][k]);
    }
    int main()
    {
        scanf("%d%d",&n,&q);
        memset(a,0,sizeof(a));
        memset(dpmin,0,sizeof(dpmin));
        memset(dpmax,0,sizeof(dpmax));
        for(int i=0; i<n; ++i)//下标0-n-1
            scanf("%d",&a[i]);
        rmq();//dp预处理
        for(int i=0; i<q; ++i)
        {
            int s,e;
            scanf("%d%d",&s,&e);
            --s,--e;
            if(s==e) printf("0\n");
            else printf("%d\n",Maxmum(s,e)-Minimum(s,e));
        }
        return 0;
    }

[USACO 2007 January Silver]: http://poj.org/searchproblem?field=source&key=USACO+2007+January+Silver