[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

 1 #include 
 2 #include
 3 #include 
 4 #include 
 5 #include 
 6 #include
 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\n");
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\n", ans);
81         }
82     }
83 
84 }

 

转载于:https://www.cnblogs.com/sainsist/p/11328824.html