poj 3468 线段树
A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 45192 | Accepted: 13257 | |
| 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
题目很好理解，就是跟新区间求和。初学线段树，一开始不会懒操作，结果每次求和都搜到树的最底层，TLE了几发。后来经KDQ大神指点，终于在知道懒操作，然后自己改改终于过了。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
struct node
{
int left,right;
long long int value;
long long int add ;
}tree[400005];
int m,n;
char ch[10];
int s,e;
long long int a,b[100005];
void push_down(int u)
{
if(tree[u].add)
{
tree[u << 1].value += (tree[u << 1].right-tree[u<<1].left+1)*tree[u].add;
tree[u<<1|1].value+=(tree[u<<1|1].right-tree[u<<1|1].left+1)*tree[u].add;
tree[u<<1].add+=tree[u].add;
tree[u<<1|1].add+=tree[u].add;
tree[u].add=0;
}
}
void create(int l,int r,int root)
{
tree[root].left=l;
tree[root].right=r;
tree[root].value=0;
tree[root].add=0;
if(l==r)
{
tree[root].value=b[l];
return;
}
int m=(tree[root].left+tree[root].right)/2;
create(l,m,root<<1);
create(m+1,r,root<<1|1);
tree[root].value=tree[root<<1].value+tree[root<<1|1].value;
}
long long int query(int l,int r,int root)
{
if(l > tree[root].right || r < tree[root].left)return 0 ;
if(tree[root].left==l&&tree[root].right==r)return tree[root].value;
else
{
push_down(root);
int m=(tree[root].left+tree[root].right)/2;
if(r<=m)return query(l,r,root<<1);
else if(l>m)return query(l,r,root<<1|1);
else return query(l,m,root<<1)+query(m+1,r,root<<1|1);
}
}
void update(int l,int r,int root,long long int v)
{
if(l>tree[root].right||r<tree[root].left)return ;
if(tree[root].left==l&&tree[root].right==r)
{
tree[root].value+=v*(tree[root].right-tree[root].left+1);
tree[root].add+=v;
return;
}
push_down(root);
int m=(tree[root].left+tree[root].right)/2;
if(r<=m)update(l,r,root<<1,v);
else if(l>m)update(l,r,root<<1|1,v);
else
{
update(l,m,root<<1,v);
update(m+1,r,root<<1|1,v);
}
tree[root].value=tree[root<<1].value+tree[root<<1|1].value;
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
scanf("%lld",&b[i]);
}
create(1,n,1);
while(m--)
{
scanf("%s",&ch);
if(ch[0]=='Q')
{
scanf("%d%d",&s,&e);
cout<<query(s,e,1)<<endl;
}
else
{
scanf("%d%d%lld",&s,&e,&a);
update(s,e,1,a);
}
}
return 0;
}