poj3468 线段树成段更新水题

poj 3468 A Simple Problem with Integers

题目：

ou 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.

题意：

就是给一个比较长的序列和两种操作，Q时询问区间所有的数字之和，C是给区间上的所有数字都加上某一个数字。

思路：

看上去就是一道线段树的成段更新，说到成段更新就一定要会打lazy标记，lazy标记的作用就是，当update操作到的某一个节点p表示的区间[l,r]完全在操作范围[x,y]内时，就不用继续操作p的子节点，直接把tree[p]+=(r-l+1)*num，lazy[p]+=num;就想当于标记好这个区间内数字的都应该加上lazy[p]的数值，就可以返回了。

然后是考虑什么时候要把lazy标记下方，就是对于当update操作，或是query操作到一点p是，发现[l,r]并不完全被[x,y]包含，这时就要继续操作他们的子节点，在操作之前就应该下方lazy标记，然后把lazy[p]清空即可，记得update操作最后要更新tree[p]=tree[p<<1]+tree[p<<1|1];

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define N 101000
using namespace std;
typedef long long ll;
ll tree[N<<2];
ll lazy[N<<2];
ll n,m;
void build(ll p,ll l,ll r)
{
if(l==r)
{
scanf("%I64d",&tree[p]);
return;
}
ll mid=(l+r)>>1;
build(p<<1,l,mid);
build((p<<1)|1,mid+1,r);
tree[p]=tree[p<<1]+tree[(p<<1)|1];
}
void pushdown(ll p,ll m)
{
if(lazy[p])
{
lazy[p<<1]+=lazy[p];
lazy[p<<1|1]+=lazy[p];
tree[p<<1]+=lazy[p]*(m-(m>>1));
tree[p<<1|1]+=lazy[p]*(m>>1);
lazy[p]=0;
}

}
void update(ll p,ll l,ll r,ll x,ll y,ll val)
{
if(x<=l&&r<=y)
{
lazy[p]+=val;
tree[p]+=val*(r-l+1);
return ;
}
pushdown(p,r-l+1);
ll mid=(l+r)>>1;
if(x<=mid)
update(p<<1,l,mid,x,y,val);
if(y>=mid+1)
update( (p<<1)|1,mid+1,r,x,y,val);
tree[p]=tree[p<<1]+tree[p<<1|1];
}
ll query(ll p,ll l,ll r,ll x,ll y)
{
if(x<=l&&r<=y)
return tree[p];
pushdown(p,r-l+1);
ll mid=(l+r)>>1;
ll tmp=0;
if(x<=mid)
tmp+=query(p<<1,l,mid,x,y);
if(y>=1+mid)
tmp+=query((p<<1)|1,mid+1,r,x,y);
return tmp;
}
int main()
{
scanf("%I64d%I64d",&n,&m);
memset(tree,0,sizeof(tree));
memset(lazy,0,sizeof(lazy));
build(1,1,n);
ll x,y,z;
for(ll i=1;i<=m;i++)
{
char s[6];
scanf("%s",s);
if(s[0]=='Q')
{
scanf("%I64d%I64d",&x,&y);
ll ans=query(1,1,n,x,y);
printf("%I64d\n",ans);
}
else
{
scanf("%I64d%I64d%I64d",&x,&y,&z);
update(1,1,n,x,y,z);
}
}
return 0;
}