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
1 #include <algorithm> 2 #include<iostream> 3 #include <cstdio> 4 #include <vector> 5 #include <cstring> 6 #include<queue> 7 using namespace std; 8 typedef long long ll; 9 const int maxn = 2e5 + 3; 10 int a[maxn], b[maxn]; 11 int root[maxn], ls[maxn * 20], rs[maxn * 20], cnt; 12 ll val[maxn * 20]; 13 void build(int l, int r, int &i) { 14 i = ++cnt; 15 val[i] = 0ll; 16 if (l == r) 17 return; 18 int mid = l + r >> 1; 19 build(l, mid, ls[i]); 20 build(mid + 1, r, rs[i]); 21 } 22 void update(int k, int l, int r, int &i) { 23 ls[++cnt] = ls[i], rs[cnt] = rs[i], val[cnt] = val[i] + 1; 24 i = cnt; 25 if (l == r) 26 return; 27 int mid = l + r >> 1; 28 if (k <= mid) 29 update(k, l, mid, ls[i]); 30 else 31 update(k, mid + 1, r, rs[i]); 32 } 33 ll query(int u, int v, int k, int l, int r) { 34 if (l == r) 35 return l; 36 int x = val[ls[v]] - val[ls[u]]; 37 int mid = l + r >> 1; 38 if (k <= x) 39 return query(ls[u], ls[v], k, l, mid); 40 else 41 return query(rs[u], rs[v], k - x, mid + 1, r); 42 } 43 int main() { 44 int n, m; 45 while (scanf("%d%d", &n, &m) != EOF) { 46 cnt = 0; 47 for (int i = 1; i <= n; i++) 48 scanf("%d", &a[i]), b[i] = a[i]; 49 sort(b + 1, b + 1 + n); 50 int k = unique(b + 1, b + 1 + n) - b - 1; 51 build(1, k, root[0]); 52 for (int i = 1; i <= n; i++) { 53 int t = lower_bound(b + 1, b + 1 + k, a[i]) - b - 1; 54 t++; 55 root[i] = root[i - 1]; 56 update(t, 1, k, root[i]); 57 } 58 for (int i = 1; i <= m; i++) { 59 int l, r; 60 scanf("%d%d", &l, &r); 61 int len = r - l + 1; 62 if (len < 3) { 63 printf("-1 "); 64 continue; 65 } 66 ll ans = -1; 67 ll x = b[query(root[l - 1], root[r], len, 1, k)]; 68 ll y = b[query(root[l - 1], root[r], len - 1, 1, k)]; 69 for (int i = 3; i <= len; i++) { 70 ll z = b[query(root[l - 1], root[r], len - i + 1, 1, k)]; 71 if (x < y + z) { 72 ans = x + y + z; 73 break; 74 } 75 else { 76 x = y; 77 y = z; 78 } 79 } 80 printf("%lld ", ans); 81 } 82 } 83 84 }