hdu 3727 Jewel(主席树学习第四弹)

题目链接：

hdu 3727

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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。

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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。
  

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AC代码：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <vector>#define MAX 200007using 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 );    } }