POJ3264——Balanced Lineup（线段树查询区间最大最小值）

Time limit

5000 ms

Case time limit

2000 ms

Memory limit

65536 kB

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.

InputLine 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

题解：

线段树建树加查询，只是树的每个节点要保存两个值（最大和最小），并且查询的时候返回结构体。

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代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;

const int MAXN = 50005;
const int INF = 0x3f3f3f3f;

int Tree[MAXN*4][2];
int Data[MAXN];

struct D{
	int MA;
	int MI;
	D(int a,int b){
		MA = a;
		MI = b;
	}
};

void Build(int temp,int left,int right){
	if(left == right){
		Tree[temp][0] = Tree[temp][1] = Data[left];
		return ;
	}
	int mid = left + (right-left)/2;
	Build(temp<<1,left,mid);
	Build(temp<<1|1,mid+1,right);
	Tree[temp][0] = max(Tree[temp<<1][0],Tree[temp<<1|1][0]);
	Tree[temp][1] = min(Tree[temp<<1][1],Tree[temp<<1|1][1]);
}

struct D query(int temp,int left,int right,int ql,int qr){
	if(ql>right || qr<left)return D(-INF,INF);
	if(ql<=left && qr>=right)return D(Tree[temp][0],Tree[temp][1]);	
	int mid = left + (right-left)/2;
	struct D a = query(temp<<1,left,mid,ql,qr);
	struct D b = query(temp<<1|1,mid+1,right,ql,qr);
	a.MA = max(a.MA,b.MA);
	a.MI = min(a.MI,b.MI);
	return a;
}

int main(){
	
	int N,M;
	scanf("%d %d",&N,&M);
	for(int i=1 ; i<=N ; i++)scanf("%d",&Data[i]);
	Build(1,1,N);
	int A,B;
	while(M--){
		scanf("%d %d",&A,&B);
		struct D t = query(1,1,N,A,B);
		printf("%d\n",t.MA-t.MI);
	}
	
	return 0;
}