luogu P3567 [POI2014] Couriers

传送门

给一个数列，每次询问一个区间内有没有一个数出现次数超过一半

区间的一半就是(r-l+1)/2

对于主席树中的两棵树L,R中的同一个节点[l,r] 如果sum的差值大于k

说明在原序列下标L`R的区间中出现了多于k个权值处于[l,r]的数字

所以按照这个套路二分就行

然后这题bzoj有双倍经验 这里粘的这道题要离散就这个了

Code:

#include<cstdio>
#include<iostream>
#include<cstring>
#include<vector>
#include<algorithm>
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
using namespace std;
typedef long long ll;
int read() {
    int as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
//head
const int N = 500005;
const int M = 10000005;
#define int ll
int n,Q,T,tot;
int a[N],b[N];
int rt[N],ls[M],rs[M],sum[M];
void ins(int l,int r,int &t,int pre,int q) {
    t = ++tot,ls[t] = ls[pre],rs[t] = rs[pre];
    sum[t] = sum[pre] + 1;
    if(l == r) return;
    int m = l+r >> 1;
    if(q <= m) ins(l,m,ls[t],ls[pre],q);
    else ins(m+1,r,rs[t],rs[pre],q);
}

int query(int l,int r,int t,int pre,int k) {
    if(l == r) return l;
    int m = l+r >> 1;
    int tmp1 = sum[ls[t]] - sum[ls[pre]];
    int tmp2 = sum[rs[t]] - sum[rs[pre]];
    if(k < tmp1) return query(l,m,ls[t],ls[pre],k);
    if(k < tmp2) return query(m+1,r,rs[t],rs[pre],k);
    return 0;
}
#undef int

int main() {
    n = read(),Q = read();
    rep(i,1,n) a[i] = b[i] = read();
    sort(b+1,b+n+1),T = unique(b+1,b+n+1) - (b+1);
    rep(i,1,n) ins(1,T,rt[i],rt[i-1],lower_bound(b+1,b+T+1,a[i]) - b);
    rep(i,1,Q) {
         ll l = read(),r = read();
        printf("%lld
",b[query(1,T,rt[r],rt[l-1],r-l+1 >> 1)]);
    }
    return 0;
}