【poj3264】Balanced Lineup

绝地灬酷狼 2022-07-18 05:47 134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;47436</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;22281</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][]

用RMQ解法相比于线段树要慢一倍，跑了1600多ms。重要的是通过此题加深理解RMQ算法，算是RMQ入门吧。

#include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    const int N = 50000+10;
    int ma[N][20],mi[N][20];
    int hei[N];
    int n,m;
    void RMQ(){
    	for(int i=1;i<=n;i++){
    		ma[i][0]=mi[i][0]=hei[i];
    	}
    	for(int j=1;j<20;j++){
    		for(int i=1;i<=n;i++){
    			if(i+(1<<j)-1<=n){
    				ma[i][j]=max(ma[i][j-1],ma[i+(1<<(j-1))][j-1]);
    				mi[i][j]=min(mi[i][j-1],mi[i+(1<<(j-1))][j-1]);
    			}
    		}
    	}
    }
    int main(){
    	while(scanf("%d%d",&n,&m)!=EOF){
    		for(int i=1;i<=n;i++){
    			scanf("%d",&hei[i]);
    		}
    		RMQ();
    		int x,y;
    		while(m--){
    			scanf("%d%d",&x,&y);
    			int k=(int)(log((double)(y-x+1))/log(2.0));//这个地方要注意，不能直接以2为低，括号的位置要正确，数据类型用浮点型。 
    			int minn,maxn;
    			minn=min(mi[x][k],mi[y-(1<<k)+1][k]);
    			maxn=max(ma[x][k],ma[y-(1<<k)+1][k]);
    			printf("%d\n",maxn-minn);
    		}
    	}
    	return 0;
    }

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