POJ 3468 A Simple Problemwith Integers(线段树:区间add,区间查询)
POJ 3468 A Simple Problemwith Integers(线段树:区间add,区间查询)
http://poj.org/problem?id=3468
题意：
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.
分析：
基本模板题目。下面代码中如果一个节点同时有addv值和sum值，那么这个节点控制元素的和只算sum，因为addv值早就已经被加了一遍了。其实我后来觉得应该两者都考虑，这样更符合思维习惯。
下面给出另一个延迟更新版的线段树区间add和区间求和的模板。
//POJ 3468 区间add，区间查询
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
//每当有add加到i节点上，不会去更新i节点的sum.
//也就是说如果要查询区间[1,n]的sum值，既要考虑sum[i]的值，也要考虑add[i]的值
const int MAXN=100000+100;
typedef long long LL;
#define lson i*2,l,m
#define rson i*2+1,m+1,r
int sum[MAXN*4];
int addv[MAXN*4];
//i节点收集下面节点的信息
void PushUp(int i,int num)
{
sum[i]=sum[i*2]+sum[i*2+1]+addv[i*2]*(num+1)/2+addv[i*2+1]*(num-(num+1)/2);
}
//将i节点的addv压下去，且更新sum[i]
void PushDown(int i,int num)
{
if(addv[i])
{
sum[i] += addv[i]*num;
addv[i*2]+=addv[i]; addv[i*2+1]+=addv[i];
addv[i]=0;
}
}
//sum[i]收集子节点的信息
//只在build线段树时会用
void PushUp(int i)
{
sum[i]=sum[i*2]+sum[i*2+1];
}
void build(int i,int l,int r)
{
addv[i]=0;
if(l==r)
{
scanf("%I64d",&sum[i]);
return ;
}
int m=(l+r)/2;
build(lson);
build(rson);
PushUp(i,r-l+1);
}
//给[ql,qr]区间加上add值
void update(int ql,int qr,int add,int i,int l,int r)
{
if(ql<=l&&r<=qr)
{
addv[i]+=add;
//sum[i] += (LL)add*(r-l+1);
return ;
}
PushDown(i,r-l+1);
int m=(l+r)/2;
if(ql<=m) update(ql,qr,add,lson);
if(m<qr) update(ql,qr,add,rson);
PushUp(i,r-l+1);
}
//查询[ql,qr]区间的sum值
int query(int ql,int qr,int i,int l,int r)
{
if(ql<=l&&r<=qr)
{
PushDown(i,r-l+1);
return sum[i];
}
PushDown(i,r-l+1);
int m=(l+r)/2;
int res=0;
if(ql<=m) res+=query(ql,qr,lson);
if(m<qr) res+=query(ql,qr,rson);
return res;
}
AC代码：1797ms
//POJ 3468 区间add，区间查询
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
//每当有add加到i节点上，直接更新i节点的sum.
//也就是说如果要查询区间[1,n]的sum值，直接sum[1]即可,不用再去考虑1的addv[1]值.
const int MAXN=100000+100;
typedef long long LL;
#define lson i*2,l,m
#define rson i*2+1,m+1,r
LL sum[MAXN*4];
LL addv[MAXN*4];
void PushDown(int i,int num)
{
if(addv[i])
{
sum[i*2] +=addv[i]*(num-(num/2));
sum[i*2+1] +=addv[i]*(num/2);
addv[i*2] +=addv[i];
addv[i*2+1]+=addv[i];
addv[i]=0;
}
}
void PushUp(int i)
{
sum[i]=sum[i*2]+sum[i*2+1];
}
void build(int i,int l,int r)
{
addv[i]=0;
if(l==r)
{
scanf("%I64d",&sum[i]);
return ;
}
int m=(l+r)/2;
build(lson);
build(rson);
PushUp(i);
}
void update(int ql,int qr,int add,int i,int l,int r)
{
if(ql<=l&&r<=qr)
{
addv[i]+=add;
sum[i] += (LL)add*(r-l+1);
return ;
}
PushDown(i,r-l+1);
int m=(l+r)/2;
if(ql<=m) update(ql,qr,add,lson);
if(m<qr) update(ql,qr,add,rson);
PushUp(i);
}
LL query(int ql,int qr,int i,int l,int r)
{
if(ql<=l&&r<=qr)
{
return sum[i];
}
PushDown(i,r-l+1);
int m=(l+r)/2;
LL res=0;
if(ql<=m) res+=query(ql,qr,lson);
if(m<qr) res+=query(ql,qr,rson);
return res;
}
int main()
{
int n,q;
while(scanf("%d%d",&n,&q)==2&&n&&q)
{
build(1,1,n);
while(q--)
{
char str[10];
scanf("%s",str);
int x,y,z;
if(str[0]=='Q')
{
scanf("%d%d",&x,&y);
printf("%I64d\n",query(x,y,1,1,n));
}
else
{
scanf("%d%d%d",&x,&y,&z);
update(x,y,z,1,1,n);
}
}
}
return 0;
}