POJ 3468 A Simple Problem with Integers【线段树区间更新入门题】

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 102132		Accepted: 31872
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

原题链接：http://poj.org/problem?id=3468

线段树区间更新入门题。区间更新用单点更新的方法去做必然会TLE，当要更新的区间正好是某个节点的区间的时候，我们就更新到此，不再继续往下更新，从而节省了时间，实现了次区间更新的高效性。

但后面更新的区间是上次更新区间的子区间的时候就要把之前保存在父节点的更新数据“下放”到对应的子区间，从而实现了操作的正确性。

对于线段树区间更新的描述网上有好多不错的博客，我就不用再去造轮子了。写法也有好多种，适合自己的才是最好的。

以下三篇代表了几种不同的写法。

此题参考博客：http://blog.csdn.net/w00w12l/article/details/7920846

http://blog.csdn.net/acdreamers/article/details/8578161

http://blog.csdn.net/acceptedxukai/article/details/6933446

AC代码：

#include <iostream>
#include <cstdio>
using namespace std;
const int maxn=100000+5;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
typedef long long LL;
LL sum[maxn<<2],add[maxn];
int a[maxn];
struct node
{
int l,r,m;
LL sum,mark;
}tree[maxn<<2];

void BuildTree(int l,int r,int k)
{
tree[k].l=l;
tree[k].r=r;
tree[k].m=(l+r)>>1;
tree[k].mark=0;
if(l==r)
{
tree[k].sum=a[l];
return;
}
int mid=(l+r)>>1;
BuildTree(l,mid,k<<1);
BuildTree(mid+1,r,k<<1|1);

tree[k].sum=(tree[k<<1].sum+tree[k<<1|1].sum);
}
void UpdateTree(int l,int r,int k,int x)
{
if(tree[k].l==l&&tree[k].r==r)
{
tree[k].mark+=x;
return;
}
tree[k].sum+=(LL)x*(r-l+1);
//int mid=(tree[k].l+tree[k].r)>>1;
int mid=tree[k].m;
if(r<=mid)
UpdateTree(l,r,k<<1,x);
else if(mid<l)
UpdateTree(l,r,k<<1|1,x);
else
{
UpdateTree(l,mid,k<<1,x);
UpdateTree(mid+1,r,k<<1|1,x);
}
}
LL QueryTree(int l,int r,int k)
{
//cout<<l<<","<<r<<","<<k<<endl;
if(tree[k].l==l&&tree[k].r==r)
return tree[k].sum+tree[k].mark*(LL)(r-l+1);

if(tree[k].mark!=0)
{
tree[k<<1].mark+=tree[k].mark;
tree[k<<1|1].mark+=tree[k].mark;

tree[k].sum+=(LL)(tree[k].r-tree[k].l+1)*tree[k].mark;//记得+1
tree[k].mark=0;
}

//int mid=(tree[k].l+tree[k].r)>>1;
int mid=tree[k].m;
if(mid>=r)
return QueryTree(l,r,k<<1);
else if(l>mid)
return QueryTree(l,r,k<<1|1);
else
return QueryTree(l,mid,k<<1)+QueryTree(mid+1,r,k<<1|1);
}
int main()
{
int n,m;
while(cin>>n>>m)
{
for(int i=1;i<=n;i++)
cin>>a[i];
BuildTree(1,n,1);
char ch[2];
int x,y,z;
while(m--)
{
scanf("%s",ch);
if(ch[0]=='Q')
{
scanf("%d%d",&x,&y);
printf("%lld\n",QueryTree(x,y,1));
}
else if(ch[0]=='C')
{
scanf("%d%d%d",&x,&y,&z);
UpdateTree(x,y,1,z);
}
}
}
return 0;
}