[POI2014]KUR-Couriers

题意

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

题解

主席树，一个数出现次数>一半，这个区间内只有这一个数满足，那么主席树直接维护所有数的出现次数，直接在树上二分查询

# include <bits/stdc++.h># define RG register# define IL inline# define Fill(a, b) memset(a, b, sizeof(a))using namespace std;typedef long long ll;const int _(500010), __(1e7);IL ll Read(){    RG char c = getchar(); RG ll x = 0, z = 1;    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);    return x * z;}int n, m, p[_], id[_], len, rt[__], ls[__], rs[__], sz[__], cnt;IL void Build(RG int &x, RG int l, RG int r){    x = ++cnt;    if(l == r) return;    RG int mid = (l + r) >> 1;    Build(ls[x], l, mid); Build(rs[x], mid + 1, r);}IL void Modify(RG int &x, RG int l, RG int r, RG int pos){    sz[++cnt] = sz[x]; ls[cnt] = ls[x]; rs[cnt] = rs[x]; sz[x = cnt]++;    if(l == r) return;    RG int mid = (l + r) >> 1;    if(pos <= mid)Modify(ls[x], l, mid, pos);    else Modify(rs[x], mid + 1, r, pos);}IL int Query(RG int A, RG int B, RG int l, RG int r, RG int half){    if(l == r) return l;    RG int sum1 = sz[ls[B]] - sz[ls[A]], sum2 = sz[rs[B]] - sz[rs[A]], mid = (l + r) >> 1;    if(sum1 > half) return Query(ls[A], ls[B], l, mid, half);    if(sum2 > half) return Query(rs[A], rs[B], mid + 1, r, half);    return 0;}int main(RG int argc, RG char* argv[]){    n = Read(); m = Read();    for(RG int i = 1; i <= n; i++) id[i] = p[i] = Read();    sort(p + 1, p + n + 1); len = unique(p + 1, p + n + 1) - p - 1;    Build(rt[0], 1, len);    for(RG int i = 1; i <= n; i++){        id[i] = lower_bound(p + 1, p + len + 1, id[i]) - p;        rt[i] = rt[i - 1];        Modify(rt[i], 1, len, id[i]);    }    for(RG int i = 1, l, r; i <= m; i++)        l = Read(), r = Read(), printf("%d\n", Query(rt[l - 1], rt[r], 1, len, (r - l + 1) >> 1));    return 0;}