[2019杭电多校第二场][hdu6601]Keen On Everything But Triangle

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6601

题意是说用给定区间内的数字组成周长最大的三角形。

大致做法就是求区间第1大，第2大和第3大然后判断是否满足，不满足再求第4大，第5大....。

原本以为复杂度爆炸，结果想想发现最坏的情况只是斐波那契的样子，每个区间也不会很大。

求区间第i大就套了个主席树

 #include <algorithm>
 #include<iostream>
 #include <cstdio>
 #include <vector>
 #include <cstring>
 #include<queue>
 using namespace std;
 typedef long long ll;
 const int maxn = 2e5 + ;
 int a[maxn], b[maxn];
 int root[maxn], ls[maxn * ], rs[maxn * ], cnt;
 ll  val[maxn * ];
 void build(int l, int r, int &i) {
     i = ++cnt;
     val[i] = 0ll;
     if (l == r)
         return;
     int mid = l + r >> ;
     build(l, mid, ls[i]);
     build(mid + , r, rs[i]);
 }
 void update(int k, int l, int r, int &i) {
     ls[++cnt] = ls[i], rs[cnt] = rs[i], val[cnt] = val[i] + ;
     i = cnt;
     if (l == r)
         return;
     int mid = l + r >> ;
     if (k <= mid)
         update(k, l, mid, ls[i]);
     else
         update(k, mid + , r, rs[i]);
 }
 ll query(int u, int v, int k, int l, int r) {
     if (l == r)
         return l;
     int x = val[ls[v]] - val[ls[u]];
     int mid = l + r >> ;
     if (k <= x)
         return query(ls[u], ls[v], k, l, mid);
     else
         return query(rs[u], rs[v], k - x, mid + , r);
 }
 int main() {
     int n, m;
     while (scanf("%d%d", &n, &m) != EOF) {
         cnt = ;
         for (int i = ; i <= n; i++)
             scanf("%d", &a[i]), b[i] = a[i];
         sort(b + , b +  + n);
         int k = unique(b + , b +  + n) - b - ;
         build(, k, root[]);
         for (int i = ; i <= n; i++) {
             int t = lower_bound(b + , b +  + k, a[i]) - b - ;
             t++;
             root[i] = root[i - ];
             update(t, , k, root[i]);
         }
         for (int i = ; i <= m; i++) {
             int l, r;
             scanf("%d%d", &l, &r);
             int len = r - l + ;
             if (len < ) {
                 printf("-1\n");
                 continue;
             }
             ll ans = -;
             ll x = b[query(root[l - ], root[r], len, , k)];
             ll y = b[query(root[l - ], root[r], len - , , k)];
             for (int i = ; i <= len; i++) {
                 ll z = b[query(root[l - ], root[r], len - i + , , k)];
                 if (x < y + z) {
                     ans = x + y + z;
                     break;
                 }
                 else {
                     x = y;
                     y = z;
                 }
             }
             printf("%lld\n", ans);
         }
     }

 }