线段树 POJ 3468 A Simple Problem with Integers

最新推荐文章于 2023-10-02 10:26:46 发布

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本文介绍了一种解决区间加查询问题的有效方法，通过构建线段树并维护懒惰标记来高效处理区间加法与区间求和操作。

这题做了我一整天，要好好记录一下！！

Description

You have N integers, A₁, A₂, ... , A_N. 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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a₊₁, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a₊₁, ... , A_b.

Output

You need to answer all Q commands in order. One answer in a line.

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

#define MAXN 100000

__int64 num[MAXN];

struct node
{
int l,r;
__int64 sum,add;
}sum[MAXN*4+5];

void build(int L,int R,int rt)
{
sum[rt].l=L;
sum[rt].r=R;
sum[rt].add=0; //记录区间的增量，如果没有增量则为0
if(L==R)
{
sum[rt].sum=num[L];
return ;
}
int m=(sum[rt].l+sum[rt].r)/2;
build(L,m,rt*2);
build(m+1,R,rt*2+1);
sum[rt].sum=sum[rt*2].sum+sum[rt*2+1].sum;
}

__int64 quary(int l,int r,int rt)
{
if(sum[rt].l>=l&&sum[rt].r<=r)
{
return sum[rt].sum;
}
int m=(sum[rt].l+sum[rt].r)/2;
if(sum[rt].add!=0) //对有增量的区间进行处理，将增量传递给自己孩子的两个节点并对两个孩子的sum进行增加，把add传递
{
sum[rt*2].sum+=(m-sum[rt].l+1)*sum[rt].add;
sum[rt*2+1].sum+=(sum[rt].r-m)*sum[rt].add;
sum[rt*2].add+=sum[rt].add;
sum[rt*2+1].add+=sum[rt].add;
sum[rt].add=0;
}
__int64 k1=0,k2=0;
if(l<=m)
k1=quary(l,r,rt*2);
if(r>m)
k2=quary(l,r,rt*2+1);
return k1+k2;
}

void updata(int l,int r,__int64 v,int rt)
{
if(sum[rt].l>=l&&sum[rt].r<=r) //对符合条件的区间进行增加，并标记增加的量
{
sum[rt].add+=v;
sum[rt].sum+=(sum[rt].r-sum[rt].l+1)*v;
return ;
}
int m=(sum[rt].l+sum[rt].r)/2;
if(sum[rt].add!=0)
{
sum[rt*2].sum+=(m-sum[rt].l+1)*sum[rt].add;
sum[rt*2+1].sum+=(sum[rt].r-m)*sum[rt].add;
sum[rt*2].add+=sum[rt].add;
sum[rt*2+1].add+=sum[rt].add;
sum[rt].add=0;
}
if(l<=m)
updata(l,r,v,rt*2);
if(r>m)
updata(l,r,v,rt*2+1);
sum[rt].sum=sum[rt*2].sum+sum[rt*2+1].sum;
}

int main(){
int i,j,k,n,m;
int a,b;
__int64 c;
char cc[2];
while( scanf("%d%d",&n,&m) != EOF)
{
for(i=1;i<=n;i++)
{
scanf("%I64d",&num[i]);
}
build(1,n,1);
while(m--){
scanf("%s",cc);
switch(cc[0])
{
case 'Q':
scanf("%d%d",&a,&b);
getchar();
printf("%I64d\n",quary(a,b,1));
break;
case 'C':
scanf("%d%d%I64d",&a,&b,&c);
getchar();
updata(a,b,c,1);
break;
}
}
}
return 0;
}