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, ... , AN. 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, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.
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;
}