hdu4441 Queue Sequence

http://www.elijahqi.win/2018/02/26/hdu4441/

Problem Description
There’s a queue obeying the first in first out rule. Each time you can either push a number into the queue (+i), or pop a number out from the queue (-i). After a series of operation, you get a sequence (e.g. +1 -1 +2 +4 -2 -4). We call this sequence a queue sequence.

Now you are given a queue sequence and asked to perform several operations:

1. insert p
   First you should find the smallest positive number (e.g. i) that does not appear in the current queue sequence, then you are asked to insert the +i at position p (position starts from 0). For -i, insert it into the right most position that result in a valid queue sequence (i.e. when encountered with element -x, the front of the queue should be exactly x).
   For example, (+1 -1 +3 +4 -3 -4) would become (+1 +2 -1 +3 +4 -2 -3 -4) after operation ‘insert 1’.
2. remove i
   Remove +i and -i from the sequence.
   For example, (+1 +2 -1 +3 +4 -2 -3 -4) would become (+1 +2 -1 +4 -2 -4) after operation ‘remove 3’.
3. query i
   Output the sum of elements between +i and -i. For example, the result of query 1, query 2, query 4 in sequence (+1 +2 -1 +4 -2 -4) is 2, 3(obtained by -1 + 4), -2 correspond.

Input
There are less than 25 test cases. Each case begins with a number indicating the number of operations n (1 ≤ n ≤ 100000). The following n lines with be ‘insert p’, ‘remove i’ or ‘query i’(0 ≤ p ≤ length (current sequence), 1 ≤ i, i is granted to be in the sequence).
In each case, the sequence is empty initially.
The input is terminated by EOF.

Output
Before each case, print a line “Case #d:” indicating the id of the test case.
After each operation, output the sum of elements between +i and -i.

Sample Input
10 insert 0 insert 1 query 1 query 2 insert 2 query 2 remove 1 remove 2 insert 2 query 3 6 insert 0 insert 0 remove 2 query 1 insert 1 query 2

Sample Output
Case #1: 2 -1 2 0 Case #2: 0 -1

Source
2012 Asia Tianjin Regional Contest
模拟队列操作 开设两个数组ta[],tb[]分别记录正数和负数在splay中的下标 每次操作的时候看一下p位置前面有多少个正数记为sz 然后 首先正数就查在p这个位置 负数怎么办找到p后面有sz个负数 插在它的前面即可 然后其他就都是正常的splay操作了 维护区间和 注意long long 相当于在splay中维护了这个区间有多少个正数有多少个负数

#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 220000
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=getchar();}
    while(ch<='9'&&ch>='0') x=x*10+ch-'0',ch=getchar();
    return x*f;
}
int sizen[N],sizep[N],ta[N],tb[N],size[N],fa[N],n,root,c[N][2],v[N],num;
ll sum[N];
inline void update(int x){
    size[x]=size[c[x][0]]+size[c[x][1]]+1;
    sizen[x]=sizen[c[x][0]]+sizen[c[x][1]]+(v[x]<0);
    sizep[x]=sizep[c[x][0]]+sizep[c[x][1]]+(v[x]>0);
    sum[x]=sum[c[x][0]]+sum[c[x][1]]+v[x];
}
inline void rotate(int x,int &tar){
    int y=fa[x],z=fa[y];
    if (y==tar) tar=x;else c[z][c[z][1]==y]=x;
    int l=c[y][1]==x,r=l^1;
    fa[c[x][r]]=y;fa[y]=x;fa[x]=z;
    c[y][l]=c[x][r];c[x][r]=y;update(y);update(x);
}
inline void splay(int x,int &tar){
    while(x!=tar){
        int y=fa[x],z=fa[y];
        if(y!=tar){
            if (c[y][0]==x^c[z][0]==y) rotate(x,tar);else rotate(y,tar);
        }rotate(x,tar);
    }
}
inline int find(int x,int sz){
    int l=c[x][0],r=c[x][1];
    if (size[l]+1==sz) return x;
    if (sz<=size[l]) return find(l,sz);else return find(r,sz-size[l]-1);
}
inline void insert1(int &x,int vv,int f){
    if (!x){x=++num;fa[x]=f;v[x]=vv;size[x]=sizep[x]=1;update(x);splay(x,root);return;}
    insert1(c[x][1],vv,x);
}
inline void insert2(int &x,int p,int vv,int f){
    if (!x){x=++num;fa[x]=f;v[x]=vv;size[x]=sizen[x]=1;update(x);splay(x,root);return;}
    int &l=c[x][0],&r=c[x][1];
    if (p<=sizen[l]+(v[x]<0)) insert2(l,p,vv,x);else insert2(r,p-(sizen[l]+(v[x]<0)),vv,x);
} 
inline void del(int x){
    splay(x,root);int pre=c[root][0],succ=c[root][1];
    while(c[pre][1]) pre=c[pre][1];while(c[succ][0]) succ=c[succ][0];
    splay(succ,root);splay(pre,c[root][0]);c[pre][1]=0;update(pre);update(root);
}
inline void init(){
    num=2;memset(size,0,sizeof(size));memset(sizen,0,sizeof(sizen));
    memset(fa,0,sizeof(fa));memset(c,0,sizeof(c));memset(sizep,0,sizeof(sizep));
    size[1]=2;v[1]=-inf;fa[2]=1;root=1;c[1][0]=2;v[2]=inf;update(2);update(1);
}
int main(){
    freopen("hdu4441.in","r",stdin);
    int test=0;
    while(~scanf("%d",&n)){printf("Case #%d:\n",++test);
        root=0;priority_queue<int,vector<int>,greater<int> >q;
        for (int i=1;i<=n;++i) q.push(i);init();
        for (int t=1;t<=n;++t){
            char op[10];scanf("%s",op);
            if (op[0]=='i'){
                int p=read(),x=q.top();q.pop();
                int xx=find(root,p+2);splay(xx,root);splay(2,c[xx][0]);int sz=sizep[c[2][1]];
                insert1(c[root][0],x,root);ta[x]=num;insert2(root,sz+1,-x,root);tb[x]=num;//print(root);puts("");
            }
            if (op[0]=='r'){
                int x=read();q.push(x);del(ta[x]);del(tb[x]);
            }
            if (op[0]=='q'){
                int x=read();splay(tb[x],root);splay(ta[x],c[root][0]);printf("%lld\n",sum[c[ta[x]][1]]);
            }
        }
    }
    return 0;
}