CF484E Sign on Fence

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;
}

来源：https://www.cnblogs.com/cjoieryl/p/8489731.html