线段树模板

#include<cstdio>
#include<algorithm>
using namespace std;
#define L o<<1
#define R (o<<1)|1
struct Node
{
int l,r,sum,Max,Min;
}Tree[1000<<2];
void PushUp(int o)
{
Tree[o].sum = Tree[o*2].sum + Tree[o*2+1].sum;
Tree[o].Max = max(Tree[o*2].Max,Tree[o*2+1].Max);
Tree[o].Min = min(Tree[o*2].Min,Tree[o*2+1].Min);
}
void Build(int o,int l,int r)
{
//首先记录l和r的值
Tree[o].l = l;
Tree[o].r = r;
if (l == r)		//到达最底层，递归终止
{
int t;
scanf ("%d",&t);		//输入数据
Tree[o].sum = Tree[o].Max = Tree[o].Min = t;		//更新节点数据
return;
}
int mid = (l+r) >> 1;		//找到中间节点
Build(o*2 , l , mid);		//递归建左子树
Build(o*2+1 , mid+1 , r);		//递归建右子树
PushUp(o);		//更新当前节点的值
}
void UpDate(int o,int l,int r,int x,int y)		//把x节点更新为y
{
if (l == r)		//递归结束
{
Tree[o].Max = Tree[o].Min = Tree[o].sum = y;		//精确找到了节点，更新
return;
}
int mid = (l+r) / 2;		//找到中间位置
if (x <= mid)
UpDate(o*2,l,mid,x,y);		//找左子树
else
UpDate(o*2+1,mid+1,r,x,y);		//找右子树
PushUp(o);		//更新当前节点
}
int QuerySum(int o,int l,int r,int x,int y)		//查找x到y的和
{
if (l == x && r == y)		//如果恰好是当前节点，就返回
{
return Tree[o].sum;
}
int mid = (l + r) / 2;
if (mid >= y)		//全在左边
return QuerySum(o*2,l,mid,x,y);
else if (x > mid)		//全在右边
return QuerySum(o*2+1,mid+1,r,x,y);
else		//一半在左一半在右
return QuerySum(o*2,l,mid,x,mid) + QuerySum(o*2+1,mid+1,r,mid+1,y);
}
int QueryMax(int o,int l,int r,int x,int y)
{
if(x == l && y == r){
return Tree[o].Max;
}
int mid = (l + r) / 2;
if(mid >= y){
return QueryMax(o*2,l,mid,x,y);
}
else if(mid < x){
return QueryMax(o*2 + 1,mid+1,r,x,y);
}
else {
return max(QueryMax(o*2,l, mid, x, mid),QueryMax( o*2 + 1, mid+1, r, mid+1, y));
}
}
int QueryMin(int o,int l,int r,int x,int y)
{
if(x == l && y == r){
return Tree[o].Min;
}
int mid = (l + r) / 2;
if(mid >= y){
return QueryMin( o*2, l, mid, x, y);
}
else if(mid < x){
return QueryMin( o*2 + 1, mid+1, r, x, y);
}
else {
return min(QueryMin( o*2, l, mid, x, mid),QueryMin( o*2 + 1, mid+1, r, mid+1, y));
}
}
int main()
{
int n;
scanf ("%d",&n);
Build(1,1,n);
//	UpDate(1,1,n,2,7);
QuerySum(1,1,n,2,4);
QueryMax(1,1,n,2,4);
QueryMin(1,1,n,2,4);
printf ("Sum:%d   Max:%d   Min:%d\n",QuerySum(1,1,n,2,4),QueryMax(1,1,n,2,4),QueryMin(1,1,n,2,4));
return 0;
}