poj3468 A Simple Problem with Integers (线段树区间最大值)

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 92127		Accepted: 28671
Case Time Limit: 2000MS

描述

You have N integers, A₁, A₂, ... , A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

输入

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁, A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

输出

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

Sample Output

4
55
9
15

Hint

The sums may exceed the range of 32-bit integers.

Source

POJ Monthly--2007.11.25, Yang Yi、

---

前几次提交的时候没有吧node[u].add清零，一直WA

第一道线段树的题，当晚没有调出来，第二天才想出来，不错

 

 

#include<cstdio>
#include<iostream>
#define LL long long
#define R(u) (u<<1|1)
#define L(u) (u<<1)
using namespace std;
const int maxx=100005;
LL a[maxx];
int n,m;
struct Node{
	int r,l;
	LL add,sum;
}node[maxx<<2];
void Pushup(int u)
{
	node[u].sum=node[L(u)].sum+node[R(u)].sum;
	return;
}
void Pushdown(int u)
{
	node[L(u)].add+=node[u].add;
	node[R(u)].add+=node[u].add;

	node[L(u)].sum+=(node[L(u)].r-node[L(u)].l+1)*node[u].add;

	node[R(u)].sum+=(node[R(u)].r-node[R(u)].l+1)*node[u].add;
	node[u].add=0;							//一定记得清零
}
void Build(int u,int left,int right)
{
	node[u].l=left,node[u].r=right;
	node[u].add=0;
	if(left==right)
	{
		node[u].sum=a[left];
		return;
	}
	int mid=(left+right)>>1;
	Build(L(u),left,mid);
	Build(R(u),mid+1,right);
	Pushup(u);
}
void update(int u,int left,int right,LL val)
{
	if(left==node[u].l&&node[u].r==right)
	{
		node[u].add+=val;
		node[u].sum+=(node[u].r-node[u].l+1)*val;
		return;
	}
	node[u].sum+=(right-left+1)*val;// 当更新区间小于这段
	int mid=(node[u].r+node[u].l)>>1;
	if(mid>=right) update(L(u),left,right,val);//在左边
	else if(mid<left) update(R(u),left,right,val);
	else {
		update(L(u),left,mid,val);
		update(R(u),mid+1,right,val);
	}
	//Pushup(u);前面已经直接算出sum后面不用再Pushup了
}
LL Qurey(int u,int left,int right)
{
	if(left==node[u].l&&node[u].r==right)
		return node[u].sum;
	if(node[u].add)Pushdown(u);
	int mid=(node[u].r+node[u].l)>>1;
	if(mid>=right) Qurey(L(u),left,right);
	else if(mid<left) Qurey(R(u),left,right);
	else return (Qurey(L(u),left,mid)+Qurey(R(u),mid+1,right));
	//Pushup(u);
}
int main()
{
	cin>>n>>m;
	LL c;
	for(int i=1;i<=n;i++)
	scanf("%I64d",a+i);
	Build(1,1,n);
	while(m--)
	{char x;int ai,an;
		scanf("%c %d %d",&x,&ai,&an);
		cin>>x>>ai>>an;
		if(x=='C')
		{
			scanf("%I64d",&c);
			update(1,ai,an,c);
		}
		else
		printf("%I64d\n",Qurey(1,ai,an));
	}
	return 0;
}