poj 3468 A Simple Problem with Integers

来源：互联网发布：北京凯恩帝数控编程编辑：程序博客网时间：2024/06/16 04:30

题目链接：http://poj.org/problem?id=3468

题目描述：给你一个序列，有两个操作，1. 给某个区间[ l, r]的每个元素加 c，2. 查询某个区间的和

解题思路：之前已经用线段树写过这道题了，现在初学splay，再用splay做一遍，其实只需要每个节点多记一个sum就可以了

//#pragma comment(linker,"/STACK:102400000,102400000")#include<stdio.h>#include<iostream>#include<string.h>#include<math.h>#include<algorithm>#include<vector>#include<map>#include<set>#include<queue>#include<string>#define ll long long#define db double#define PB push_back#define lson k<<1#define rson k<<1|1using namespace std;const int N = 100005;struct node *null;struct node{    node *ch[2];    int v,lz,s;    ll sum;    int cmp(int x)    {        int d=x-ch[0]->s;        if(d==1) return -1;        return d<=0?0:1;    }    void add(node *p,int val)    {        if(p==null) return;        p->lz+=val,p->sum+=(ll)val*(ll)p->s,p->v+=val;    }    void pushdown()    {        if(lz!=0)        {            add(ch[0],lz),add(ch[1],lz);            lz=0;        }    }    void pushup()    {        sum=v+ch[0]->sum+ch[1]->sum;        s=1+ch[0]->s+ch[1]->s;    }} seq[N],*root;int a[N];node* build(int l,int r){    if(l>r) return null;    int mid=(l+r)>>1;    node *tmp=build(l,mid-1);    node *p=&seq[mid];    p->lz=0,p->v=a[mid],p->ch[0]=tmp;    p->ch[1]=build(mid+1,r);    p->pushup();    return p;}void rotate(node *&o,int d){    node *k=o->ch[d^1];o->ch[d^1]=k->ch[d],k->ch[d]=o;    o->pushup();    o=k;}void splay(node *&o,int k){    int d=o->cmp(k);    o->pushdown();    if(d!=-1)    {        if(d==1) k-=o->ch[0]->s+1;        node *p=o->ch[d];        int d2=p->cmp(k);        if(d2==1) k-=p->ch[0]->s+1;        p->pushdown();        if(d2!=-1)        {            splay(p->ch[d2],k);            if(d==d2) rotate(o,d^1);            else rotate(o->ch[d],d);        }        rotate(o,d^1);    }    o->pushup();}int main(){#ifdef PKWV    freopen("in.in","r",stdin);#endif // PKWV    int n,Q;    scanf("%d%d",&n,&Q);    for(int i=1;i<=n;i++) scanf("%d",&a[i+1]);    null=new node();    null->s=0,null->sum=0LL;    root=build(1,n+2);    while(Q--)    {        char ch[10];        int a,b,c;        scanf("%s%d%d",ch,&a,&b);        splay(root,a);        splay(root->ch[1],b-a+2);        node *tmp=root->ch[1]->ch[0];        if(ch[0]=='Q')        {            printf("%I64d\n",tmp->sum);        }else        {            scanf("%d",&c);            tmp->lz+=c;            tmp->v+=c;            tmp->sum+=(ll)c*tmp->s;        }    }    return 0;}

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