POJ 3264 Balanced Lineup（线段树）

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 23699		Accepted: 11019
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

Source

USACO 2007 January Silver

 

 

 

这题早就做过了。。。

重新做一次，认真学一下线段树吧！

/*
POJ 3264 Balanced Lineup
线段树
求区间中最大数和最小数的差
*/
#include<stdio.h>
#include<algorithm>
#include<iostream>
using namespace std;
const int MAXN=50010;
const int INF=0x3fffffff;
int nMax,nMin;
struct Node
{
    int l,r;
    int nMin,nMax;
}segTree[3*MAXN];
void Build(int i,int l,int r)
{
    segTree[i].l=l;
    segTree[i].r=r;
    if(l==r)
    {
        scanf("%d",&segTree[i].nMin);
        segTree[i].nMax=segTree[i].nMin;
        return;
    }
    int mid=((l+r)>>1);
    Build(i<<1,l,mid);
    Build((i<<1)|1,mid+1,r);
    segTree[i].nMin=min(segTree[i<<1].nMin,segTree[(i<<1)|1].nMin);
    segTree[i].nMax=max(segTree[i<<1].nMax,segTree[(i<<1)|1].nMax);
}
void Query(int i,int l,int r)
{
    if(segTree[i].nMax<=nMax&&segTree[i].nMin>=nMin)return;
    if(segTree[i].l==l&&segTree[i].r==r)
    {
        nMin=min(segTree[i].nMin,nMin);
        nMax=max(segTree[i].nMax,nMax);
        return;
    }
    int mid=(segTree[i].l+segTree[i].r)>>1;
    if(r<=mid)Query(i<<1,l,r);
    else if(l>mid)Query((i<<1)|1,l,r);
    else
    {
        Query(i<<1,l,mid);
        Query((i<<1)|1,mid+1,r);
    }
}
int main()
{
    int n,q;
    int l,r;
    while(scanf("%d%d",&n,&q)!=EOF)
    {
        Build(1,1,n);
        while(q--)
        {
            scanf("%d%d",&l,&r);
            nMax=-INF;
            nMin=INF;
            Query(1,l,r);
            printf("%d\n",nMax-nMin);
        }
    }
    return 0;
}