[POI 2014] Couriers

[题目链接]

          https://www.lydsy.com/JudgeOnline/problem.php?id=3524

[算法]

         首先离线 ， 将询问按右端点排序

         如果我们知道[l , r]这个区间中[L , mid]中的数有多少个和[mid + 1 , R]中的数有多少个 ， 则可以通过二分的方式求出答案

         可持久化线段树可以完成这个任务

         时间复杂度 : O(NlogN)

[代码] 

          

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 5e5 + 10;

struct query
{
        int l , r;
        int id;
} q[MAXN];

int n , m , idx;
int a[MAXN] , lson[MAXN << 5] , rson[MAXN << 5] , sum[MAXN << 5] , root[MAXN] , tmp[MAXN] , ans[MAXN];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
inline bool cmp(query a , query b)
{
        return a.r < b.r;
}
inline void build(int &k , int l , int r)
{
        k = ++idx;
        if (l == r) return;
        int mid = (l + r) >> 1;
        build(lson[k] , l , mid);
        build(rson[k] , mid + 1 , r);        
}
inline void modify(int &k , int old , int l , int r , int pos , int value)
{
        k = ++idx;
        lson[k] = lson[old] , rson[k] = rson[old];
        sum[k] = sum[old] + value;
        if (l == r) return;
        int mid = (l + r) >> 1;
        if (mid >= pos) modify(lson[k] , lson[k] , l , mid , pos , value);
        else modify(rson[k] , rson[k] , mid + 1 , r , pos , value);        
}
inline int query(int rt1 , int rt2 , int l , int r , int x)
{
        if (l == r) return l;
        int mid = (l + r) >> 1;
        if (sum[lson[rt1]] - sum[lson[rt2]] > x) return query(lson[rt1] , lson[rt2] , l , mid , x);
        else if (sum[rson[rt1]] - sum[rson[rt2]] > x) return query(rson[rt1] , rson[rt2] , mid + 1 , r , x);
        else return 0;
} 

int main()
{
        
        read(n); read(m);
        for (int i = 1; i <= n; i++) 
        {
                read(a[i]);
                tmp[i] = a[i];
        }
        sort(tmp + 1 , tmp + n + 1);
        int len = unique(tmp + 1 , tmp + n + 1) - tmp - 1;
        for (int i = 1; i <= n; i++) a[i] = lower_bound(tmp + 1 , tmp + len + 1 , a[i]) - tmp;
        for (int i = 1; i <= m; i++)
        {
                read(q[i].l);
                read(q[i].r);        
                q[i].id = i;
        }
        sort(q + 1 , q + m + 1 , cmp);
        build(root[0] , 1 , len);
        int now = 1;
        for (int i = 1; i <= n; i++)
        {
                modify(root[i] , root[i - 1] , 1 , n , a[i] , 1);
                while (now <= m && q[now].r == i) ans[q[now].id] = tmp[query(root[i] , root[q[now].l - 1] , 1 , n , (q[now].r - q[now].l + 1) / 2)] , ++now;
        }
        for (int i = 1; i <= m; i++) printf("%d\n" , ans[i]);
        
        return 0;
    
}