当前位置: 代码迷 >> 综合 >> POJ3264Balanced Lineup 线段树
  详细解决方案

POJ3264Balanced Lineup 线段树

热度:83   发布时间:2023-11-25 14:19:07.0

Balanced Lineup
Time Limit: 5000MS   Memory Limit: 65536K
Total Submissions: 62002   Accepted: 28961
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

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

Sample Output

6
3
0

AC代码:

//#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const int MAXSIZE=50000+10;
const int inf=0x3f3f3f3f;
struct node{int value,value1;int rchild,lchild;
}tree[MAXSIZE*4];
int father[MAXSIZE];
int n,q;
void BuiltTree(int i,int l,int r){tree[i].lchild=l;tree[i].rchild=r;tree[i].value=-inf;tree[i].value1=inf;if(l==r){father[l]=i;return;}BuiltTree(i<<1,l,(l+r)/2);BuiltTree((i<<1)+1,(l+r)/2+1,r);
}
void Update(int i){if(i<=1) return;int fi=i>>1;int a=tree[fi<<1].value;int b=tree[(fi<<1)+1].value;int a1=tree[fi<<1].value1;int b1=tree[(fi<<1)+1].value1;tree[fi].value=max(a,b);tree[fi].value1=min(a1,b1);Update(i>>1);
}
int MAX,MIN;
void Query(int i,int l,int r){if(tree[i].lchild==l&&tree[i].rchild==r){MAX=max(MAX,tree[i].value);MIN=min(MIN,tree[i].value1);return;}i=i<<1;if(l<=tree[i].rchild){if(r<=tree[i].rchild) Query(i,l,r);else Query(i,l,tree[i].rchild);}i++;if(r>=tree[i].lchild){if(l>=tree[i].lchild) Query(i,l,r);else Query(i,tree[i].lchild,r);}
}
int main(){std::ios::sync_with_stdio(false);cin>>n>>q;BuiltTree(1,1,n);for(int i=1;i<=n;i++){cin>>tree[father[i]].value;tree[father[i]].value1=tree[father[i]].value;Update(father[i]);}int a,b;while(q--){cin>>a>>b;MAX=-inf;MIN=inf;Query(1,a,b);cout<<MAX-MIN<<endl;}return 0;
}