POJ3264Balanced Lineup（经典线段树）

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 39618		Accepted: 18600
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

题意：

给定Q (1 ≤ Q ≤ 200,000)个数A1,A2 … AQ,， 多次求任一区间Ai – Aj中最大数和最小数的 差。 

思路：

经典的线段树问题，写题的时候一定要细心。我把一个*打成+ 错了好几次。。

#include<iostream>
#include<stdio.h>
#include<string.h>
#define max(a,b) a>b?a:b
#define min(a,b) a<b?a:b
using namespace std;
const int INF = 0xffffff0; 
int minV=INF;
int maxV=-INF;
struct Node
{
	int L,R;
	int minV,maxV;
	int Mid()
	{
		return (L+R)/2;
	}
};
 Node tree[800010];
 void BuildTree(int root,int L,int R)
 {
 	tree[root].L=L;
 	tree[root].R=R;
 	tree[root].minV=INF;
 	tree[root].maxV=-INF;
 	if(L!=R)
 	{
	 	BuildTree(2*root+1,L,(L+R)/2);
	 	BuildTree(2*root+2,(L+R)/2+1,R);
	 }
 }
void Insert(int root,int i,int v)
 {
 	if(tree[root].L==tree[root].R)
 	{
	 	tree[root].minV=tree[root].maxV=v;
	 	return;
	 }
	 tree[root].minV=min(tree[root].minV,v);
	 tree[root].maxV=max(tree[root].maxV,v);
	 if(i<=tree[root].Mid())
	 	Insert(2*root+1,i,v);
 	else Insert(2*root+2,i,v);
 }
void Query(int root,int s,int e)
 {
 	if(tree[root].minV>=minV&&tree[root].maxV<=maxV)
 		return;
	if(tree[root].L==s&&tree[root].R==e)
	{
		minV=min(minV,tree[root].minV);
		maxV=max(maxV,tree[root].maxV);
		return;
	}
	if(e<=tree[root].Mid())
		Query(root*2+1,s,e);
	else if(s>tree[root].Mid())
		Query(root*2+2,s,e);
	else{
		Query(2*root+1,s,tree[root].Mid());
		Query(2*root+2,tree[root].Mid()+1,e);
	}
 }
 int main()
 {
 	int n,q,h;
 	int i,j,k;
 	scanf("%d%d",&n,&q);
 	BuildTree(0,1,n);
	 for(i=1;i<=n;i++)
 	{
	 	scanf("%d",&h);
	 	Insert(0,i,h);
	 }
	 for(i=0;i<q;i++)
	 {
 		int s,e;
 		scanf("%d%d",&s,&e);
 		minV=INF;
 		maxV=-INF;
 		Query(0,s,e);
 		printf("%d\n",maxV-minV);
 	}
	 return 0; 
 }