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| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 37869 | Accepted: 17751 | |
| 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
Output
Sample Input
6 3 1 7 3 4 2 5 1 5 4 6 2 2
Sample Output
6 3 0
解题思路：先求最大值，再求最小值
/*******************************************************************************
* Author : jinbao
* Email : dongjinbao913106840144@gmail.com
* Last modified : 2015-05-11 20:30
* Filename : poj3264.cpp
* Description :
* *****************************************************************************/
#include <iostream>
#include <stdio.h>
using namespace std;
#define min(x,y) x<y?x:y
#define max(x,y) x>y?x:y
const int MAX = 50000+1;
struct cow{
int Min,Max;
}p[4*MAX];
int a[MAX];
void build(int node,int begin,int end){
if (begin==end){
scanf("%d",&a[begin]);
p[node].Min=p[node].Max=a[begin];
}
else{
build(node*2,begin,(begin+end)/2);
build(node*2+1,(begin+end)/2+1,end);
p[node].Min=min(p[node*2].Min,p[node*2+1].Min);
p[node].Max=max(p[node*2].Max,p[node*2+1].Max);
}
}
int query(int node,int begin,int end,int left,int right,int flag){
if (end<left || begin>right){
if (flag==0)
return 0xffffff;
return -1;
}
if (begin>=left && end<=right){
if (flag==0)
return p[node].Min;
return p[node].Max;
}
int m=query(2*node,begin,(begin+end)/2,left,right,flag);
int n=query(2*node+1,(begin+end)/2+1,end,left,right,flag);
if (flag==0)
return min(m,n);
return max(m,n);
}
int main(){
int n,q,l,r,Min,Max;
while (~scanf("%d%d",&n,&q)){
build(1,1,n);
while (q--){
scanf("%d%d",&l,&r);
int ans = query(1,1,n,l,r,1) - query(1,1,n,l,r,0);
printf("%d\n",ans);
}
}
return 0;
}
poj3264 Balanced Lineup（求区间的最大值与最小值之差）
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原文地址：http://blog.csdn.net/codeforcer/article/details/45896071
