poj 3264 【线段树】

此题为入门级线段树

题意：给定Q（1<=Q<=200000）个数A1A2…AQ，多次求任一区间Ai-Aj中最大数和最小数的差

#include<algorithm>
#include<cstdio>
#include<string>
#include<string.h>
#include<iostream>
using namespace std;
typedef long long LL;
const int INF = 0x7FFFFFFF;
const int maxn = 1e3 + 10;

int minV = INF;
int maxV = -INF;
struct Node
{
	int L, R;
	int minV, maxV;
	//Node *pLeft, *pRight;
	int Mid()
	{
		return (L + R) / 2;
	}
};
Node tree[800010];//4倍叶子节点的数量就够

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)
//将第i个数，其值为v，插入线段树
{
	if (tree[root].L == tree[root].R)
	{
		//成立则亦有tree[root].R==i
		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)
//查询区间[s,e]中的最小值和最大值，如果更优就记在全局变量里
{
	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(2 * root + 1, s, e);
	else if (s > tree[root].Mid())
		Query(2 * root + 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;
}