hdu 4699 Editor(双向链表+随便维护前缀和)

Editor

Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1532    Accepted Submission(s): 480

Problem Description

 

Sample Input

8I 2I -1I 1Q 3LDRQ 2

 

Sample Output

23
Hint
The following diagram shows the status of sequence after each instruction:
 

 

Source

2013 Multi-University Training Contest 10

 

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 题意：

就是模仿编辑器。只是只能编辑数字。然后询问是询问光标前的数列的前k项和的最大值。

思路：

非常像Splay但是完全没那么复杂。用链表模拟。然后随便用什么维护前i项和就行了。比赛时没多想就用的线段树。其实用个数组就够了，不过懒得写了，

详细见代码：

#include<algorithm>#include<iostream>#include<string.h>#include<stdio.h>using namespace std;const int INF=0x3f3f3f3f;const int maxn=1000010;typedef long long ll;#define lson L,mid,ls#define rson mid+1,R,rsint mav[maxn<<2],st[maxn];struct node{    int pre,next,rk,val,sum;} pos[maxn];void build(int L,int R,int rt){    mav[rt]=-INF;    if(L==R)        return;    int ls=rt<<1,rs=ls|1,mid=(L+R)>>1;    build(lson);    build(rson);}void update(int L,int R,int rt,int p,int d){    if(L==R)    {        mav[rt]=d;        return;    }    int ls=rt<<1,rs=ls|1,mid=(L+R)>>1;    if(p<=mid)        update(lson,p,d);    else        update(rson,p,d);    mav[rt]=max(mav[ls],mav[rs]);}int qu(int L,int R,int rt,int l,int r){    if(l<=L&&R<=r)        return mav[rt];    int ls=rt<<1,rs=ls|1,mid=(L+R)>>1,tp=-INF;    if(l<=mid)        tp=max(tp,qu(lson,l,r));    if(r>mid)        tp=max(tp,qu(rson,l,r));    return tp;}int main(){    int q,i,x,ps,ns,tot;    char cmd[10];    while(~scanf("%d",&q))    {        tot=0;        build(1,q,1);        for(i=q+5;i>=0;i--)            st[tot++]=i;        ps=st[--tot];        pos[ps].pre=pos[ps].next=-1;        pos[ps].rk=pos[ps].sum=0;        for(i=1;i<=q;i++)        {            scanf("%s",cmd);            if(cmd[0]=='I')            {                scanf("%d",&x);                ns=st[--tot];                pos[ns].val=x;                pos[ns].rk=pos[ps].rk+1;                pos[ns].sum=pos[ps].sum+x;                pos[ns].pre=ps;                pos[ns].next=pos[ps].next;                if(pos[ps].next!=-1)                    pos[pos[ps].next].pre=ns;                pos[ps].next=ns;                update(1,q,1,pos[ns].rk,pos[ns].sum);                ps=ns;            }            else if(cmd[0]=='D')            {                if(pos[ps].pre==-1)                    continue;                pos[pos[ps].pre].next=pos[ps].next;                if(pos[ps].next!=-1)                    pos[pos[ps].next].pre=pos[ps].pre;                st[tot++]=ps;                ps=pos[ps].pre;            }            else if(cmd[0]=='L')            {                if(pos[ps].pre!=-1)                    ps=pos[ps].pre;            }            else if(cmd[0]=='R')            {                if(pos[ps].next==-1)                    continue;                pos[pos[ps].next].rk=pos[ps].rk+1;                pos[pos[ps].next].sum=pos[ps].sum+pos[pos[ps].next].val;                ps=pos[ps].next;                update(1,q,1,pos[ps].rk,pos[ps].sum);            }            else            {                scanf("%d",&x);                printf("%d\n",qu(1,q,1,1,x));            }            //printf("ps %d\n",pos[ps].rk);        }    }    return 0;}