BZOJ 3524

传送门

题目分析

标准主席树，按照位置插入每个数，对于询问l, r, 在l-1,r两树上按照线段树搜索次数大于(r - l + 1) / 2的数。

code

#include<bits/stdc++.h>
using namespace std;

const int N = 500050, M = 500050;
int n, m;
int a[N];

struct node{
	int lc, rc, cnt;
}tr[N * 25];
int pool, rt[N];

namespace IO{
	inline int read(){
	    int i = 0, f = 1; char ch = getchar();
	    for(; (ch < '0' || ch > '9') && ch != '-'; ch = getchar());
	    if(ch == '-') f = -1, ch = getchar();
	    for(; ch >= '0' && ch <= '9'; ch = getchar()) i = (i << 3) + (i << 1) + (ch -'0');
	    return i * f;
	}

	inline void wr(int x){
		if(x < 0) putchar('-'), x = -x;
		if(x > 9) wr(x / 10);
		putchar(x % 10 + '0');
	}
}using namespace IO;

inline void insert(int x, int &y, int l, int r, int v){
	tr[y = ++pool] = tr[x];
	tr[y].cnt++;
	if(l == r) return;
	int mid = l + r >> 1;
	if(v <= mid) insert(tr[x].lc, tr[y].lc, l, mid, v);
	else insert(tr[x].rc, tr[y].rc, mid + 1, r, v);
}

inline int query(int nl, int nr, int l, int r, int x){
//	if(tr[nr].cnt - tr[nl].cnt < x) return 0;
	if(l == r) return l;
	int mid = l + r >> 1;
	if(tr[tr[nr].lc].cnt - tr[tr[nl].lc].cnt >= x) return query(tr[nl].lc, tr[nr].lc, l, mid, x);
	else if(tr[tr[nr].rc].cnt - tr[tr[nl].rc].cnt >= x) return query(tr[nl].rc, tr[nr].rc, mid + 1, r, x);
	else return 0;
}

int main(){
	freopen("h.in", "r", stdin);
	n = read(), m = read();
	for(int i = 1; i <= n; i++) a[i] = read(), insert(rt[i - 1], rt[i], 1, 500000, a[i]);
	for(int i = 1; i <= m; i++){
		int l = read(), r = read(), len = (r - l + 1) / 2;
		int ans = query(rt[l - 1], rt[r], 1, 500000, len + 1);
		wr(ans), putchar('
');
	}
	return 0;
}