P3567 [POI2014]KUR-Couriers 主席树

Posted 2021-01-29 dukelv

这个题比一般主席树还要简单，但是用来练习主席树再好不过了，在这里我再放一下主席树板子。

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<ctime>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
#define duke(i,a,n) for(register int i = a;i <= n;i++)
#define lv(i,a,n) for(register int i = a;i >= n;i--)
#define clean(a) memset(a,0,sizeof(a))
const int INF = 1 << 30;
typedef long long ll;
typedef double db;
template <class T>
void read(T &x)
{
    char c;
    bool op = 0;
    while(c = getchar(), c < ‘0‘ || c > ‘9‘)
        if(c == ‘-‘) op = 1;
    x = c - ‘0‘;
    while(c = getchar(), c >= ‘0‘ && c <= ‘9‘)
        x = x * 10 + c - ‘0‘;
    if(op) x = -x;
}
template <class T>
void write(T x)
{
    if(x < 0) putchar(‘-‘), x = -x;
    if(x >= 10) write(x / 10);
    putchar(‘0‘ + x % 10);
}
const int N = 2e5 + 5;
struct node
{
    int l,r,sum;
}tree[N * 20];
int root[N];
int cnt = 0,n,m;
void insert(int lst,int &now,int l,int r,int pos)
{
    tree[++cnt] = tree[lst];
    now = cnt;
    ++tree[now].sum;
    if(l == r) return;
    int mid = (l + r) >> 1;
    if(pos <= mid) insert(tree[lst].l,tree[now].l,l,mid,pos);
    else insert(tree[lst].r,tree[now].r,mid + 1,r,pos);
}
int query(int lst,int now,int l,int r,int val)
{
    if(l == r) return tree[now].sum > val ? l : 0;;
    int tmp = tree[tree[now].l].sum - tree[tree[lst].l].sum;
//    cout<<lst<<" "<<now<<" "<<l<<" "<<r<<endl;
    int mid = (l + r) >> 1;
    if(tmp > val) return query(tree[lst].l,tree[now].l,l,mid,val);
    else if(tree[tree[now].r].sum - tree[tree[lst].r].sum > val)
    return query(tree[lst].r,tree[now].r,mid + 1,r,val);
    return 0;
}
int a,tot;
int main()
{
    read(n);read(m);
    duke(i,1,n)
    {
        read(a);
        insert(root[i - 1],root[i],1,n,a);
    }
    /*for(int i = 1;i <= 20;i++)
    {
        printf("%d %d %d
",tree[i].l,tree[i].r,tree[i].sum);
    }*/
    while(m--)
    {
        int x,y;
        read(x);read(y);
        printf("%d
",query(root[x - 1],root[y],1,n,(y - x + 1) >> 1));
    }
    return 0;
}