题目链接：

hdu 3727

题目大意：

给出n个操作，插入一个数，或者查询：
- 查询区间第k大
- 查询某个数的排名
- 查询整体第k大
最后问三种查询的结果之和。

题目分析：

  • 首先对数进行离散化
  • 然后对数字建立主席树，第k大的做法详见我的主席树学习第一弹
  • 查询某个数的排名，树状数组直接做就好了，记录每个点是否出现过，然后找到sum(x)就是前缀一共有多少个数，也就是当前数的排名。

AC代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#define MAX 200007

using namespace std;

typedef long long LL;

int root[MAX],cnt_trees,n,len,h[MAX],hcnt;
LL ans1,ans2,ans3;

struct Query
{
    int f,s,t,k;
    void init ( int a , int b  ,int c , int d )
    {f=a,s=b,t=c,k=d;}
}q[MAX<<1];

struct Tree
{
    int l,r;
    int sum;
}tree[4000007];

void build ( int& u , int l , int r )
{
    u = cnt_trees++;
    tree[u].sum = 0;
    if ( l == r ) return;
    int mid = l+r>>1;
    build ( tree[u].l , l , mid );
    build ( tree[u].r , mid+1 , r );
}

void update ( int p , int& u , int l , int r , int x  )
{
    u = cnt_trees++;
    tree[u] = tree[p];
    tree[u].sum++;
    if ( l == r ) return;
    int mid = l+r>>1;
    if ( x > mid )
        update ( tree[p].r , tree[u].r , mid+1 , r , x );
    else 
        update ( tree[p].l , tree[u].l , l , mid , x );
}

int query ( int p , int u , int l , int r , int k )
{
    int sum = tree[tree[u].l].sum - tree[tree[p].l].sum;
    if ( l == r ) return l;
    int mid = l+r>>1;
    if ( sum >= k ) return query ( tree[p].l , tree[u].l , l , mid , k );
    else return query ( tree[p].r , tree[u].r , mid+1 , r , k-sum );
}

int c[MAX<<1];

int lowbit ( int x )
{
    return x&-x;
}

void add ( int x )
{
    while ( x < hcnt )
    {
        c[x]++;
        x += lowbit ( x );
    }
}

int sum ( int x )
{
    int ret = 0;
    while ( x )
    {
        ret += c[x];
        x -= lowbit ( x );
    }
    return ret;
}

char s[30];

int main ( )
{
    int cc = 1;
    while ( ~scanf ( "%d" , &n ) )
    {
        cnt_trees = 0;
        len = hcnt = 1;
        memset ( c , 0 , sizeof ( c ) );
        for ( int i = 1 ; i <= n ; i++ )
        {
            int x,y,z;
            scanf ( "%s" , s );
            if ( s[0] == 'I' )
            {
                q[i].f = 0;
                scanf ( "%d" , &q[i].k );
                h[hcnt++] = q[i].k;
            }
            else
            {
                int id = s[6]-48;
                if ( id == 1 )
                    scanf ( "%d%d%d" , &x , &y , &z );
                else 
                    scanf ( "%d" , &z );
                q[i].init ( id , x , y , z );
            }
        }
        sort ( h+1 , h+hcnt );
        build ( root[0] , 1 , hcnt-1 );
        ans1 = ans2 = ans3 = 0;
        for ( int i = 1 ; i <= n ; i++ )
        {
            if ( q[i].f == 0 )
            {
                int x = lower_bound ( h , h+hcnt , q[i].k )-h;
                update ( root[len-1] , root[len] , 1 , hcnt-1 , x );
                add ( x );
                len++;
            }
            else if ( q[i].f == 1 )
                ans1 += h[query ( root[q[i].s-1] , root[q[i].t] , 1 , hcnt-1 , q[i].k )];
            else if ( q[i].f == 2 )
            {
                int x = lower_bound ( h , h+hcnt , q[i].k )-h;
                ans2 += sum(x);
            }
            else if ( q[i].f == 3 )
                ans3 += h[query ( root[0] , root[len-1] , 1 , hcnt-1 , q[i].k )];
        }
        printf ( "Case %d:\n" , cc++ );
        printf ( "%I64d\n%I64d\n%I64d\n" , ans1 , ans2 , ans3 );
    } 
}