线段树的lazy（poj3468）

2023-08-08 11:25:28

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 73163		Accepted: 22585
Case Time Limit: 2000MS

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.

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

Hint

The sums may exceed the range of 32-bit integers.

Source

POJ Monthly--2007.11.25, Yang Yi

　　最裸的线段树lazy标记

 #include <iostream>

 #include <cstdio>

 using namespace std;

 const int N = ;

 typedef  long long LL;

 LL sum[N<<]; //sum用来存储每个节点的子节点数值的总和

 LL add[N<<];//add用来记录该节点的每个数值应该加多少

 struct Node{

     int l,r;//表示改点的左右区间

     int mid(){//结构体函数

         return (l+r)>>;

     }

 } tree[N<<];

 void PushUp(int rt){//算某一节点的左右孩子值的和

     sum[rt] = sum[rt<<] + sum[rt<<|];

 }

 void PushDown(int rt,int m){//rt当前节点  m此节点的区间长度

     if(add[rt]!=){//如果当前节点lazy标记不为 0

         add[rt<<] += add[rt];//左右孩子lazy累加

         add[rt<<|] += add[rt];

         sum[rt<<] += add[rt]*(m-(m>>));//更新计算左右孩子

         sum[rt<<|] += add[rt] * (m>>);

         add[rt] = ;//小细节，但很重要，不能忘记lazy用过后清零

     }

 }

 void build(int l,int r,int rt){

     tree[rt].l = l;

     tree[rt].r = r;

     add[rt] = ;//lazy标记记为0

     if(l == r){

         scanf("%lld",&sum[rt]);//把子节点信息记录在sum里

         return ;

     }

     int m = tree[rt].mid();

     build(l,m,rt<<);//建立左右孩子，这时编号是按照类似“宽搜”来编的

     build(m+,r,rt<<|);

     PushUp(rt);//算出此节点的权值

 }

 void update(int c,int l,int r,int rt){//当前节点rt，在l和r区间上加上c

     if(l<=tree[rt].l&&tree[rt].r<=r){//当前节点所表示的区间完全被所要更新的区间包含

         add[rt]+=c;//lazy累加,add[i]数组是针对i左右孩子的

         sum[rt]+=(LL)c*(tree[rt].r-tree[rt].l+);//这句话很重要和下面的PushUP()类似

         return;

     }

     PushDown(rt,tree[rt].r - tree[rt].l + );//用rt的lazy更新其子节点

     int m = tree[rt].mid();

     if(l<=m) update(c,l,r,rt<<);

     if(m+<=r) update(c,l,r,rt<<|);

     PushUp(rt);

 }

 LL query(int l,int r,int rt){//当前节点为rt，求l，到r的和

     if(l<=tree[rt].l&&tree[rt].r<=r){//区间完全包含，可以直接返回

         return sum[rt];

     }

     //因为此时用到rt节点，所以才更新lazy

     PushDown(rt,tree[rt].r-tree[rt].l + );

     int m = tree[rt].mid();

     LL res = ;

     if(l<= m)//要求的区间有一部分在mid的左边

         res += query(l,r,rt<<);

     if(m+<=r)

         res += query(l,r,rt<<|);

     return res;

 }

 int main(){

     int n,m;

     scanf("%d %d",&n,&m);

     build(,n,);

     while(m--){

     char ch[];

     scanf("%s",ch);

     int a,b,c;

     if(ch[] == 'Q'){

         scanf("%d %d", &a,&b);

         printf("%lld\n",query(a,b,));

     }

     else{

         scanf("%d %d %d",&a,&b,&c);

         cin>>a>>b>>c;

         update(c,a,b,);

         }

     }

     return ;

 }

还有一种较好理解的方法：

 #include<iostream>

 #include<cstdio>

 #include<cstdlib>

 #include<cmath>

 #include<algorithm>

 #include<cstring>

 using namespace std;

 typedef long long LL;

 inline LL read(){

     LL x=,f=;char ch=getchar();

     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

     return x*f;

 }

 const LL maxn=;

 struct node{

     LL num;

     LL l,r;

     LL lc,rc;

     LL sum,lazy;

 }seg[maxn*];

 LL a[maxn];

 LL N,M,tot=;

 inline void updata_son(LL root){

     LL d=seg[root].lazy;

     if(d!=){

         LL lson=seg[root].lc; LL rson=seg[root].rc;

         seg[lson].lazy+=d; seg[rson].lazy+=d; seg[root].lazy=;

         seg[lson].sum+=d*(seg[lson].r-seg[lson].l+);

         seg[rson].sum+=d*(seg[rson].r-seg[rson].l+);

     }

 }

 inline void Build(LL root,LL l,LL r){

     seg[root].num=root;

     seg[root].l=l; seg[root].r=r;

     if(l==r){

         seg[root].sum=a[l];

         return ;

     }

     LL mid=(l+r)>>;

     seg[root].lc=++tot; Build(tot,l,mid);

     seg[root].rc=++tot; Build(tot,mid+,r);

     seg[root].sum=seg[seg[root].lc].sum+seg[seg[root].rc].sum;

 }

 inline LL query(LL root,LL l,LL r){

     if(l<=seg[root].l&&seg[root].r<=r){

         return seg[root].sum;

     }

     updata_son(root);

     LL ans=;

     LL mid=(seg[root].l+seg[root].r)>>;

     if(l<=mid) ans+=query(seg[root].lc,l,r);

     if(mid+<=r) ans+=query(seg[root].rc,l,r);

     return ans;

 }

 inline void change(LL root,LL l,LL r,LL delta){

     if(l<=seg[root].l&&seg[root].r<=r){

         seg[root].sum+=(seg[root].r-seg[root].l+)*delta;

         seg[root].lazy+=delta;

         return ;

     }

     updata_son(root);

     LL mid=(seg[root].l+seg[root].r)>>;

     if(l<=mid) change(seg[root].lc,l,r,delta);

     if(mid+<=r) change(seg[root].rc,l,r,delta);

     seg[root].sum=seg[seg[root].lc].sum+seg[seg[root].rc].sum;

 }

 int main(){

     N=read(); M=read();

     for(LL i=;i<=N;i++) a[i]=read();

     Build(,,N);

     LL ll,rr,de;

     char x[];

     for(LL i=;i<=M;i++){

         scanf("%s",x);

         if(x[]=='C'){

             ll=read(); rr=read(); de=read();

             change(,ll,rr,de);

             continue;

         }

         if(x[]=='Q'){

             ll=read(); rr=read();

             printf("%lld\n",query(,ll,rr));

         }

     }

     return ;

 }

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