2019 Multi-University Training Contest 2

主席树

用主席数来维护区间第k大，然后不能构成三角形的最小条件就是斐波那契数列。

题目给的范围只有1e9，最坏情况下也只有44项斐波那契数列，也就是说我们从第1大，第2大，第3大依次往下判断，也最多遍历44次，肯定可以在规定时间内求解。

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define FAST_IO ios::sync_with_stdio(false)
using namespace std;
typedef long long LL;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int ret = 0, w = 0; char ch = 0;
    while(!isdigit(ch)){
        w |= ch == '-', ch = getchar();
    }
    while(isdigit(ch)){
        ret = (ret << 3) + (ret << 1) + (ch ^ 48);
        ch = getchar();
    }
    return w ? -ret : ret;
}
inline int lcm(int a, int b){ return a / __gcd(a, b) * b; }
template <typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}

const int N = 200005;
int n, m, tot, k, a[N], b[N], tree[N*20], lc[N*20], rc[N*20], root[N];

void init(){
    full(a, 0), full(b, 0);
    full(tree, 0), full(root, 0);
    full(lc, 0), full(rc, 0);
    k = tot = 0;
}

int buildTree(int l, int r){
    int cur = ++tot;
    if(l == r) return cur;
    int mid = (l + r) >> 1;
    lc[cur] = buildTree(l, mid);
    rc[cur] = buildTree(mid + 1, r);
    return cur;
}

int insert(int rt, int l, int r, int p){
    int cur = ++tot;
    tree[cur] = tree[rt] + 1, lc[cur] = lc[rt], rc[cur] = rc[rt];
    if(l == r) return cur;
    int mid = (l + r) >> 1;
    if(p <= mid) lc[cur] = insert(lc[rt], l, mid, p);
    else rc[cur] = insert(rc[rt], mid + 1, r, p);
    return cur;
}

int query(int a, int b, int l, int r, int k){
    int p = tree[lc[b]] - tree[lc[a]];
    if(l == r) return l;
    int mid = (l + r) >> 1;
    if(k <= p) return query(lc[a], lc[b], l, mid, k);
    else return query(rc[a], rc[b], mid + 1, r, k - p);
}

int main(){

    while(~scanf("%d%d", &n, &m)){
        init();
        for(int i = 1; i <= n; i ++){
            scanf("%d", &a[i]);
            b[i] = a[i];
        }
        sort(b + 1, b + n + 1);
        k = unique(b + 1, b + n + 1) - b - 1;
        root[0] = buildTree(1, k);
        for(int i = 1; i <= n; i ++){
            int p = lower_bound(b + 1, b + k + 1, a[i]) - b;
            root[i] = insert(root[i - 1], 1, k, p);
        }
        int l = 0, r = 0;
        while(m --){
            scanf("%d%d", &l, &r);
            int cnt = r - l + 1;
            if(cnt < 3){
                printf("-1
");
                continue;
            }
            bool good = false;
            int fst = b[query(root[l - 1], root[r], 1, k, cnt)];
            int sed = b[query(root[l - 1], root[r], 1, k, cnt - 1)];
            for(int i = cnt - 2; i >= 1; i --){
                int val = b[query(root[l - 1], root[r], 1, k, i)];
                if(1LL * val + sed > fst){
                    printf("%lld
", 1LL * val + sed + fst);
                    good = true;
                    break;
                }
                fst = sed, sed = val;
            }
            if(!good) printf("-1
");
        }
    }
    return 0;
}