CDQZ data structure exercises (Update)

challenge 0：

Water problems, violence can be modified too.

#include<bits/stdc++.h>
using namespace std;
const int maxn=1000010;
int n,m,k,p;
char flag;
int a[maxn];
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
    }
    for(int i=1;i<=m;i++)
    {
        cin>>flag;
        if(flag=='Q')
        {
            scanf("%d",&p);
            printf("%d\n",a[p]);
        }
        else
        {
            scanf("%d%d",&k,&p);
            a[k]=p;        
        }
    }
    return 0;
}

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challenge 1：

Query a version certainly is persistent data structure, a single point of modification, with persistence segment tree can be a time to open a new chain is modified, thus greatly reducing the overhead space.

Query a version, from the root of a particular operation can query down.

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+7;
int a[maxn];
int n,m;
int cnt;
struct node{
    int lson,rson,val;//左儿子，右儿子，权值 
}tree[maxn*4];
int root[maxn]; 
char opt;
int x,y;
int tim;
void copynode(int neww,int old){
    tree[neww].lson=tree[old].lson;
    tree[neww].rson=tree[old].rson;
    tree[neww].val=tree[old].val;
}
void build(int rt,int l,int r){
    if(l==r){
        tree[rt].val=a[l];
        return;
    }
    tree[rt].lson=++cnt;//动态开点 
    tree[rt].rson=++cnt;
    int mid=l+r>>1;
    build(tree[rt].lson,l,mid);
    build(tree[rt].rson,mid+1,r);
}
void update(int rt,int l,int r,int ori,int pos,int x){
    copynode(rt,ori);
    if(l==r){
        tree[rt].val=x;
        return;
    }
    int mid=l+r>>1;
    if(pos<=mid){
        tree[rt].lson=++cnt;
        update(tree[rt].lson,l,mid,tree[ori].lson,pos,x);
    } 
    else{
        tree[rt].rson=++cnt;
        update(tree[rt].rson,mid+1,r,tree[ori].rson,pos,x);
    }
}
int query(int rt,int l,int r,int pos){
    if(l==r) return tree[rt].val;
    int mid=(l+r)>>1;
    if(pos<=mid) return query(tree[rt].lson,l,mid,pos);
    else return query(tree[rt].rson,mid+1,r,pos);
}
int main(){
    scanf("%d%d",&n,&m);
    root[0]=++cnt;
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    build(root[0],1,n);
    while(m--){
        cin>>opt>>x>>y;
        if(opt=='Q'){
            tim++;
            root[tim]=root[tim-1];
            printf("%d\n",query(root[y],1,n,x));
        }
        if(opt=='M' ) { 
            Tim ++ ; 
            the root [Tim] = CNT ++; // new node to open 
            Update (the root [Tim], . 1 , n-, the root [tim- . 1 ], X, Y); 
        } 
    } 
    return  0 ; 
}

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challenge 3：

Tree line single point of modification.

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+7;
struct node{
    int l,r;
    long long sum;
}tree[maxn*20];
int a[maxn],n,m;
int x,v,l1,r1;
char opt[5];
void build(int now,int l,int r){
    tree[now].l=l,tree[now].r=r;
    if(l==r){
        tree[now].sum=a[l];
        return;
    }
    int mid=(l+r)>>1;
    build(now<<1,l,mid);
    build(now<<1|1,mid+1,r);
    tree[now].sum=tree[now<<1].sum+tree[now<<1|1].sum;
}
void update(int now,int x,int v){
    if(tree[now].l==tree[now].r){
        tree[now].sum=v;
        return;
    } 
    int mid=(tree[now].l+tree[now].r)>>1;
    if(x<=mid) update(now<<1,x,v);
    else update(now<<1|1,x,v);
    tree[now].sum=tree[now<<1].sum+tree[now<<1|1].sum;
}
long long query(int now,int l,int r){
    if(tree[now].l>=l&&tree[now].r<=r) return tree[now].sum;
    long long val=0;
    int mid=(tree[now].l+tree[now].r)>>1;
    if(l<=mid) val+=query(now<<1,l,r);
    if(r>mid) val+=query(now<<1|1,l,r);
    return val;
}
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    build(1,1,n);
    for(int i=1;i<=m;i++){
        scanf("%s",opt);
        if(opt[0]=='M'){
            scanf("%d%d",&x,&v);
            update(1,x,v);
        }
        else{
            scanf("%d%d",&l1,&r1);
            printf("%d\n",query(1,l1,r1));
        }
    }
    return 0;
} 

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challenge 5：

Tree line interval assignment.

Warning the FBI : This title toxic data, there will be negative when the assignment, note labeled lazy initialization.

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+7;
struct node{
    int l,r;
    long long lazy;
    long long sum;
}tree[maxn*25];
int n,m;
long long a[maxn];
int l,r;
long long v;
char opt[666];
void build(int now,int l,int r){
    tree[now].l=l,tree[now].r=r,tree[now].lazy=-99999999;
    if(l==r){
        tree[now].sum=a[l];
        return;
    }
    int mid=(l+r)>>1;
    build(now<<1,l,mid);
    build(now<<1|1,mid+1,r);
    tree[now].sum=tree[now<<1].sum+tree[now<<1|1].sum;
}
void pushdown(int now){
    if(tree[now].lazy!=-99999999){
        tree[now<<1].sum=(tree[now<<1].r-tree[now<<1].l+1)*tree[now].lazy;
        tree[now<<1|1].sum=(tree[now<<1|1].r-tree[now<<1|1].l+1)*tree[now].lazy;
        tree[now<<1].lazy=tree[now].lazy;
        tree[now<<1|1].lazy=tree[now].lazy;
        tree[now].lazy=-99999999;
    }
}
void update(int now,int l,int r,long long v){
    if(tree[now].l>=l&&tree[now].r<=r){
        tree[now].sum=(tree[now].r-tree[now].l+1)*v;
        tree[now].lazy=v;
        return;
    }
    pushdown(now);
    int mid=(tree[now].l+tree[now].r)>>1;
    if(l<=mid) update(now<<1,l,r,v);
    if(r>mid) update(now<<1|1,l,r,v);
    tree[now].sum=tree[now<<1].sum+tree[now<<1|1].sum;
}
long long query(int now,int l,int r){
    if(tree[now].l>=l&&tree[now].r<=r) return tree[now].sum;
    pushdown(now);
    int mid=(tree[now].l+tree[now].r)>>1;
    long long val=0;
    if(l<=mid) val+=query(now<<1,l,r);
    if(r>mid) val+=query(now<<1|1,l,r);
    return val;
}
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
    build(1,1,n); 
    for(int i=1;i<=m;i++){
        scanf("%s",opt);
        if(opt[0]=='M') scanf("%d%d%lld",&l,&r,&v),update(1,l,r,v);
        else scanf("%d%d",&l,&r),printf("%lld\n",query(1,l,r));
    }
    return 0;
}

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