POJ 3468 A Simple Problem with Integers //线段树的成段更新

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 59046		Accepted: 17974
Case Time Limit: 2000MS

Description

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.

Input

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.

Output

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

/*
	区间更新的lazy操作。
*/
#include <stdio.h>

struct node
{
	int l, r;
	__int64 sum;
	__int64 lazy;	//当成段更新时，往往不用更新到单个的点。lazy操作大大节省了时间。
}tree[300005];
int h[100005];
__int64 sum;		//int超限

void build(int l, int r, int n)
{
	int mid;
	tree[n].l = l;
	tree[n].r = r;
	tree[n].lazy = 0;	//赋初值
	if(l==r)
	{
		tree[n].sum = h[l];
		return ;
	}
	mid = (l+r)/2;
	build(l, mid, 2*n);
	build(mid+1, r, 2*n+1);
	
	tree[n].sum = tree[2*n].sum+tree[2*n+1].sum;
}
void add(int l, int r, __int64 k, int n)
{
	int mid;
	if(tree[n].l==l && tree[n].r==r)	//当须要更新的段 与 结点相应的段吻合时，直接把此结点的lazy值更新就可以，不须要再向下更新。
	{
		tree[n].lazy += k;
		return;
	}
	
	tree[n].sum += k*(r-l+1);		//当此区间包括须要更新的区间，但不吻合时，须要向下继续查找，此时须要更新这个父节点的sum值。
	
	mid = (tree[n].l + tree[n].r)/2;
	if(r <= mid)
		add(l, r, k, 2*n);
	else if(l >=mid+1)
		add(l, r, k, 2*n+1);
	else
	{
		add(l, mid, k, 2*n);
		add(mid+1, r, k, 2*n+1);
	}
}
void qu(int l, int r, int n)
{
	int mid;
	if(tree[n].l==l && tree[n].r==r)
	{
		sum += tree[n].sum + (r-l+1)*tree[n].lazy;	//当查找的段与 此结点的段吻合时，sum 值等于这个结点的sum加上lazy乘区间长度的值。
		return ;
	}
	
	if(tree[n].lazy!=0 && tree[n].l!=tree[n].r)	//当查找区间为此结点相应区间的子集时，须要将此结点相应的lazy值下放到其子节点，并把此结点的lazy值置为0。
	{
		add(tree[2*n].l, tree[2*n].r, tree[n].lazy, n);
		add(tree[2*n+1].l, tree[2*n+1].r, tree[n].lazy, n);
		tree[n].lazy = 0;
	}
	mid = (tree[n].l + tree[n].r)/2;

	if(l >= mid+1)
		qu(l, r, 2*n+1);
	else if(r <= mid)
		qu(l, r, 2*n);
	else
	{
		qu(l, mid, 2*n);
		qu(mid+1, r, 2*n+1);
	}
}

int main()
{
	int n, q;
	int i;
	int a, b, c;
	char ch[10];
	
	scanf("%d%d", &n, &q);
	for(i=1; i<=n; i++)
		scanf("%d", &h[i]);
	
	build(1, n, 1);
	while(q--)
	{
		scanf("%s", ch);
		if(ch[0]=='Q')
		{
			scanf("%d%d", &a, &b);
			sum = 0;
			qu(a, b, 1);
			printf("%I64d\n", sum);
		}
		else
		{
			scanf("%d%d%d", &a, &b, &c);
			add(a, b, c, 1);
		}
	}
	return 0;
}