數組 stream hdu nes eve resp help first tails

Problem Description

Jimmy wants to make a special necklace for his girlfriend. He bought many beads with various sizes, and no two beads are with the same size. Jimmy can‘t remember all the details about the beads, for the necklace is so long. So he turns to you for help.

Initially, there is no bead at all, that is, there is an empty chain. Jimmy always sticks the new bead to the right of the chain, to make the chain longer and longer. We number the leftmost bead as Position 1, and the bead to its right as Position 2, and so on. Jimmy usually asks questions about the beads‘ positions, size ranks and actual sizes. Specifically speaking, there are 4 kinds of operations you should process:

Insert x
Put a bead with size x to the right of the chain (0 < x < 231, and x is different from all the sizes of beads currently in the chain)
Query_1 s t k
Query the k-th smallest bead between position s and t, inclusive. You can assume 1 <= s <= t <= L, (L is the length of the current chain), and 1 <= k <= min (100, t-s+1)
Query_2 x
Query the rank of the bead with size x, if we sort all the current beads by ascent order of sizes. The result should between 1 and L (L is the length of the current chain)
Query_3 k
Query the size of the k-th smallest bead currently (1 <= k <= L, L is the length of the current chain)

Input

There are several test cases in the input. The first line for each test case is an integer N, indicating the number of operations. Then N lines follow, each line contains one operation, as described above.

You can assume the amount of "Insert" operation is no more than 100000, and the amounts of "Query_1", "Query_2" and "Query_3" are all less than 35000.
There are several test cases in the input. The first line for each test case is an integer N, indicating the number of operations. Then N lines follow, each line contains one operation, as described above.

You can assume the amount of "Insert" operation is no more than 100000, and the amounts of "Query_1", "Query_2" and "Query_3" are all less than 35000.Query the rank of the bead with size x, if we sort all the current beads by ascent order of sizes. The result should between 1 and L (L is the length of the current chain)
Query_3 k
Query the size of the k-th smallest bead currently (1 <= k <= L, L is the length of the current chain)

Output

Output 4 lines for each test case. The first line is "Case T:", where T is the id of the case. The next 3 lines indicate the sum of results for Query_1, Query_2 and Query_3, respectively.

Sample Input

10 Insert 1 Insert 4 Insert 2 Insert 5 Insert 6 Query_11 5 5 Query_1 2 3 2 Query_2 4 Query_3 3 Query_3 1

Sample Output

Case 1: 10 3 5

Hint

The answers for the 5 queries are 6, 4, 3, 4, 1, respectively.

題意：

輸入n，再輸入n個操作，操作有四種

0， Insert X：插入x到序列末尾
1， query_1 L R X：在當前序列中的[l,r]區間找第x小的數。
2， query_2 X：在當前序列中，輸出X是第幾小的數。
3， query_3 X：找到當前序列中第X小的數是幾。
然後輸出的是3種query的和。

思路：

操作1和操作3是查詢區間第k小，主席樹。

操作2是求排名，用樹狀數組或者二分。

在線轉化為離線。

求Kth的主席樹的模板見我之前寫的：http://www.cnblogs.com/hua-dong/p/7931778.html

操作2為二分的代碼：

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<map> 
using namespace std;
const int maxn=200010;
int a[maxn],b[maxn],Innum,n;//a是位置，b是值。Innum是輸入的個數，即 
int cnt,ql,qr,Case=0; 
long long sum1,sum2,sum3;
struct questions
{
    int opt;//0,1,2,3
    int x,y,k;
}Qst[maxn<<1];
struct PLTree
{
    int ch[maxn * 20][2],sum[maxn * 20],rt[maxn];
    void build(int& now,int l,int r)
    {
        now = ++ cnt;
        sum[now] = 0;
        if(l == r) return ;
        int Mid = (l + r)>>1;
        build(ch[now][0],l,Mid);
        build(ch[now][1],Mid + 1,r);
    }
    void insert(int& now,int last,int l,int r,int pos)
    {
        now = ++ cnt;
        ch[now][0]=ch[last][0];
        ch[now][1]=ch[last][1];
        sum[now] = sum[last] + 1;
        if(l == r) return ; 
        int Mid = (l+r) >> 1;
        if(pos <= Mid) insert(ch[now][0],ch[last][0],l,Mid,pos);
        else insert(ch[now][1],ch[last][1],Mid + 1,r,pos);
    }
    int query(int ss,int tt,int l,int r,int k)
    {
         if(l == r) return l;
         int Mid =(l + r) >> 1,tmp = sum[ch[tt][0]] - sum[ch[ss][0]];
         if(k <= tmp) return query(ch[ss][0],ch[tt][0],l,Mid,k);
         else return query(ch[ss][1],ch[tt][1],Mid + 1,r,k - tmp);
    }
};
PLTree P;
void _init()
{
    Innum=cnt=0;
    sum1=sum2=sum3=0;
}
void _scanf()
{
    char chr[10];
    for(int i=1;i<=n;i++){
        scanf("%s",chr);
        if(chr[0]==‘I‘){
            Qst[i].opt=0;
            scanf("%d",&Qst[i].x);
            a[++Innum]=Qst[i].x;
            b[Innum]=a[Innum];
        }
        else {   
            Qst[i].opt=chr[6]-‘0‘;
            if(chr[6]==‘1‘)   scanf("%d%d%d",&Qst[i].x,&Qst[i].y,&Qst[i].k);
            else   scanf("%d",&Qst[i].k);
        }
    }
}
void _disp()
{
    sort(b+1,b+Innum+1);
    for(int i=1;i<=Innum;i++)
      a[i]=lower_bound(b+1,b+Innum+1,a[i])-b;
}
void _work()
{
    P.build(P.rt[0],1,Innum);
    int nowcnt=0,ans;
    for(int i=1;i<=n;i++){
        if(Qst[i].opt==0) {
            nowcnt++;
            P.insert(P.rt[nowcnt],P.rt[nowcnt-1],1,Innum,a[nowcnt]); 
        }
        else if(Qst[i].opt==1){
            ans=P.query(P.rt[Qst[i].x-1],P.rt[Qst[i].y],1,Innum,Qst[i].k);
            sum1+=b[ans];
        }
        else if(Qst[i].opt==2){
            int L=1,R=nowcnt;
            while(L<=R){
                int Mid=(L+R)>>1;
                ans=P.query(P.rt[0],P.rt[nowcnt],1,Innum,Mid);
                if(b[ans]==Qst[i].k) {
                   sum2+=Mid;
                   break;
                }
                else if(b[ans]<Qst[i].k) L=Mid+1;
                else R=Mid-1;
            }   
        }
        else {
            ans=P.query(P.rt[0],P.rt[nowcnt],1,Innum,Qst[i].k);
            sum3+=b[ans];
        }
    }
    printf("Case %d:\n",++Case);
    printf("%lld\n%lld\n%lld\n",sum1,sum2,sum3);
}
int main()
{
    while(~scanf("%d",&n)){        
        _init();
        _scanf();//輸入 
        _disp();//離散
        _work();
    }
    return 0;
}

操作2為樹狀數組的代碼：

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<map> 
using namespace std;
const int maxn=200010;
int a[maxn],b[maxn],Innum,n;//a是位置，b是值。Innum是操作為輸入的個數，即 
int cnt,ql,qr,Case=0; 
long long  sum1,sum2,sum3;
int c[maxn<<1];
int lowbit(int x )
{
    return x&-x;
}
void add( int x )
{
    while(x < Innum){
        c[x]++;
        x += lowbit ( x );
    }
}
int sum(int x )
{
    int ret = 0;
    while (x){
        ret += c[x];
        x -= lowbit ( x );
    }
    return ret;
}
struct questions
{
    int opt;//0,1,2,3
    int x,y,k;
}Qst[maxn<<1];
struct PLTree
{
    int ch[maxn * 20][2],sum[maxn * 20],rt[maxn];
    void build(int& now,int l,int r)
    {
        now = ++ cnt;
        sum[now] = 0;
        if(l == r) return ;
        int Mid = (l + r)>>1;
        build(ch[now][0],l,Mid);
        build(ch[now][1],Mid + 1,r);
    }
    void insert(int& now,int last,int l,int r,int pos)
    {
        now = ++ cnt;
        ch[now][0]=ch[last][0]; 
        ch[now][1]=ch[last][1];
        sum[now] = sum[last] + 1;
        if(l == r) return ; 
        int Mid = (l+r) >> 1;
        if(pos <= Mid) insert(ch[now][0],ch[last][0],l,Mid,pos);
        else insert(ch[now][1],ch[last][1],Mid + 1,r,pos);
    }
    int query(int ss,int tt,int l,int r,int k)
    {
         if(l == r) return l;
         int Mid =(l + r) >> 1,tmp = sum[ch[tt][0]] - sum[ch[ss][0]];
         if(k <= tmp) return query(ch[ss][0],ch[tt][0],l,Mid,k);
         else return query(ch[ss][1],ch[tt][1],Mid + 1,r,k - tmp);
    }
};
PLTree P;
void _init()
{
    Innum=cnt=0;
    sum1=sum2=sum3=0;
    memset(c,0,sizeof(c));
}
void _scanf()
{
    char chr[10];
    for(int i=1;i<=n;i++){
        scanf("%s",chr);
        if(chr[0]==‘I‘){
            Qst[i].opt=0;
            scanf("%d",&Qst[i].x);
            a[++Innum]=Qst[i].x;
            b[Innum]=a[Innum];
        }
        else {   
            Qst[i].opt=chr[6]-‘0‘;
            if(chr[6]==‘1‘)   scanf("%d%d%d",&Qst[i].x,&Qst[i].y,&Qst[i].k);
            else   scanf("%d",&Qst[i].k);
        }
    }
}
void _disp()
{
    sort(b+1,b+Innum+1);
    for(int i=1;i<=Innum;i++)
      a[i]=lower_bound(b+1,b+Innum+1,a[i])-b;
}
void _work()
{
    P.build(P.rt[0],1,Innum);
    int nowcnt=0,ans;
    for(int i=1;i<=n;i++){
        if(Qst[i].opt==0) {
            nowcnt++;
            P.insert(P.rt[nowcnt],P.rt[nowcnt-1],1,Innum,a[nowcnt]); 
            add(a[nowcnt]);
        }
        else if(Qst[i].opt==1){
            ans=P.query(P.rt[Qst[i].x-1],P.rt[Qst[i].y],1,Innum,Qst[i].k);
            sum1+=b[ans];
        }
        else if(Qst[i].opt==2){
            int tmp=sum(lower_bound(b+1,b+Innum+1,Qst[i].k)-b);
            sum2+=tmp;
        }
        else {
            ans=P.query(P.rt[0],P.rt[nowcnt],1,Innum,Qst[i].k);
            sum3+=b[ans];
        }
    }
    printf("Case %d:\n",++Case);
    printf("%lld\n%lld\n%lld\n",sum1,sum2,sum3);
}
int main()
{
    while(~scanf("%d",&n)){        
        _init();
        _scanf();//輸入 
        _disp();//離散
        _work();
    }
    return 0;
}

HDU3727 Jewel（主席樹+樹狀數組（或二分））