题意
给定一个长度为n的数列，有m次询问，询问形如l r k
要你在区间[l,r]内选一个长度为k的区间，求区间最小数的最大值
Sol
二分答案
怎么判定，每种数字开一棵线段树
某个位置上的数大于等于它为1
那么就是求区间最大的1的序列长度大于k
二分的最优答案一定在这个区间内，否则不优
排序后就是用主席树优化空间
之前\(build\)一下，因为区间有长度不好赋值
# 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 _(1e5 + 5);
const int __(2e6 + 5);
IL int Input(){
RG int x = 0, z = 1; RG char c = getchar();
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, tot, rt[_], a[_], id[_], ls[__], rs[__], o[_], len;
struct Data{
int maxl, maxr, maxn, len;
IL void Init(){
maxl = maxr = maxn = len = 0;
}
} T[__], Ans;
IL int Cmp(RG int x, RG int y){
return a[x] < a[y];
}
IL Data Merge(RG Data A, RG Data B){
RG Data ret;
ret.maxl = A.maxl, ret.maxr = B.maxr, ret.len = A.len + B.len;
ret.maxn = max(A.maxr + B.maxl, max(A.maxn, B.maxn));
if(A.maxl == A.len) ret.maxl = A.len + B.maxl;
if(B.maxr == B.len) ret.maxr = B.len + A.maxr;
return ret;
}
IL void Modify(RG int &x, RG int l, RG int r, RG int p){
ls[++tot] = ls[x], rs[tot] = rs[x], T[tot] = T[x], x = tot;
if(l == r){
T[x].maxl = T[x].maxr = T[x].maxn = 1;
return;
}
RG int mid = (l + r) >> 1;
if(p <= mid) Modify(ls[x], l, mid, p);
else Modify(rs[x], mid + 1, r, p);
T[x] = Merge(T[ls[x]], T[rs[x]]);
}
IL void Query(RG int x, RG int l, RG int r, RG int L, RG int R){
if(L <= l && R >= r){
Ans = Merge(Ans, T[x]);
return;
}
RG int mid = (l + r) >> 1;
if(L <= mid) Query(ls[x], l, mid, L, R);
if(R > mid) Query(rs[x], mid + 1, r, L, R);
}
IL void Build(RG int &x, RG int l, RG int r){
T[x = ++tot].len = r - l + 1;
if(l == r) return;
RG int mid = (l + r) >> 1;
Build(ls[x], l, mid), Build(rs[x], mid + 1, r);
}
int main(RG int argc, RG char* argv[]){
n = Input();
for(RG int i = 1; i <= n; ++i) id[i] = i, o[i] = a[i] = Input();
sort(id + 1, id + n + 1, Cmp), sort(o + 1, o + n + 1);
len = unique(o + 1, o + n + 1) - o - 1;
Build(rt[len + 1], 1, n);
for(RG int i = len, j = n; i; --i){
rt[i] = rt[i + 1];
for(; j && a[id[j]] == o[i]; --j)
Modify(rt[i], 1, n, id[j]);
}
m = Input();
for(RG int i = 1; i <= m; ++i){
RG int l = Input(), r = Input(), k = Input();
RG int L = 1, R = len, ans = 0;
while(L <= R){
RG int mid = (L + R) >> 1;
Ans.Init();
Query(rt[mid], 1, n, l, r);
if(Ans.maxn >= k) ans = mid, L = mid + 1;
else R = mid - 1;
}
printf("%d\n", o[ans]);
}
return 0;
}