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poj 2104主席树求区间第k小

时间:2017-08-24 00:07:47      阅读:126      评论:0      收藏:0      [点我收藏+]

标签:pair   tps   comm   .net   ace   str   fixed   ack   define   

POJ - 2104

题意:求区间第k小

思路:无修改主席树

AC代码:

#include "iostream"
#include "iomanip"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#pragma comment(linker, "/STACK:102400000,102400000")
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a,x) memset(a,x,sizeof(a))
#define step(x) fixed<< setprecision(x)<<
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define ll long long
#define endl ("\n")
#define ft first
#define sd second
#define lrt (rt<<1)
#define rrt (rt<<1|1)
using namespace std;
const ll mod=1e9+7;
const ll INF = 1e18+1LL;
const int inf = 1e9+1e8;
const double PI=acos(-1.0);
const int N=1e5+100;

int ls[N*20],rs[N*20],rt[N],sum[N*20],cnt;
int n,m,ran[N];

void creat(int &cur, int l, int r){
    cur=++cnt;
    sum[cur]=0;
    if(l==r) return;
    int mid=l+r>>1;
    creat(ls[cur], l, mid);
    creat(rs[cur], mid+1, r);
}

void update(int &cur, int l, int r, int p, int last){
    cur=++cnt;
    ls[cur]=ls[last];
    rs[cur]=rs[last];
    sum[cur] = sum[last]+1;
    if(l==r) return;
    int mid=l+r>>1;
    if(p<=mid) update(ls[cur], l, mid, p, ls[cur]);
    else update(rs[cur], mid+1, r, p, rs[cur]);
}

int query(int cur_l, int cur_r, int l, int r, int k){
    if(l==r) return l;
    int t=sum[ls[cur_r]]-sum[ls[cur_l]];
    int mid=l+r>>1;
    if(t>=k) return query(ls[cur_l], ls[cur_r], l, mid, k);
    else return query(rs[cur_l], rs[cur_r], mid+1, r, k-t);
}

struct Node{
    int id, x;
    bool friend operator< (Node a, Node b){
        return a.x<b.x;
    }
    bool friend operator== (Node a, Node b){
        return a.x==b.x;
    }
}a[N];
int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    while(cin>>n>>m){
        for(int i=1; i<=n; ++i){
            cin>>a[i].x; a[i].id=i;
        }
        sort(a+1,a+1+n); cnt=0;
        int sz=n;//unique(a+1,a+1+n) - (a+1); ///cout<<sz<<endl;
        for(int i=1; i<=sz; ++i) ran[a[i].id]=i;
        creat(rt[0], 1, n);
        for(int i=1; i<=n; ++i){ //cout<<ran[i]<<endl;
            update(rt[i], 1, n, ran[i], rt[i-1]);
        }
        while(m--){
            int l,r,k; cin>>l>>r>>k;
            cout<<a[query(rt[l-1], rt[r], 1, sz, k)].x<<endl;
        }
    }
    return 0;
}

 

poj 2104主席树求区间第k小

标签:pair   tps   comm   .net   ace   str   fixed   ack   define   

原文地址:http://www.cnblogs.com/max88888888/p/7420702.html

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