|
| 1 | +/* |
| 2 | + Idea: |
| 3 | + - Segment tree. |
| 4 | + - Try each element with the maximum element after it |
| 5 | + and count the number of elements between them less |
| 6 | + than the current element. |
| 7 | +*/ |
| 8 | + |
| 9 | +#include <bits/stdc++.h> |
| 10 | + |
| 11 | +using namespace std; |
| 12 | + |
| 13 | +int const N = 1e5 + 1; |
| 14 | +int n, a[N], mx[N]; |
| 15 | +vector<int> seg[4 * N]; |
| 16 | + |
| 17 | +void build(int at, int l, int r) { |
| 18 | + if(l == r) { |
| 19 | + seg[at].push_back(a[l]); |
| 20 | + return; |
| 21 | + } |
| 22 | + |
| 23 | + int mid = (l + r) >> 1; |
| 24 | + build(at << 1, l, mid); |
| 25 | + build(at << 1 | 1, mid + 1, r); |
| 26 | + |
| 27 | + int i = 0, j = 0; |
| 28 | + for(; i < seg[at << 1].size() && j < seg[at << 1 | 1].size();) |
| 29 | + if(seg[at << 1][i] < seg[at << 1 | 1][j]) |
| 30 | + seg[at].push_back(seg[at << 1][i++]); |
| 31 | + else |
| 32 | + seg[at].push_back(seg[at << 1 | 1][j++]); |
| 33 | + while(i < seg[at << 1].size()) |
| 34 | + seg[at].push_back(seg[at << 1][i++]); |
| 35 | + while(j < seg[at << 1 | 1].size()) |
| 36 | + seg[at].push_back(seg[at << 1 | 1][j++]); |
| 37 | +} |
| 38 | + |
| 39 | +int s, e, tar; |
| 40 | +int get(int at, int l, int r) { |
| 41 | + if(l > e || r < s) |
| 42 | + return 0; |
| 43 | + |
| 44 | + if(l >= s && r <= e) |
| 45 | + return upper_bound(seg[at].begin(), seg[at].end(), tar) - seg[at].begin(); |
| 46 | + |
| 47 | + int mid = (l + r) >> 1; |
| 48 | + return get(at << 1, l, mid) + get(at << 1 | 1, mid + 1, r); |
| 49 | +} |
| 50 | + |
| 51 | +int solve() { |
| 52 | + memset(mx, -1, sizeof mx); |
| 53 | + for(int i = n, cur = 1; i > 0; --i) |
| 54 | + while(cur <= a[i]) |
| 55 | + mx[cur++] = i; |
| 56 | + |
| 57 | + for(int i = 0; i < 4 * N; ++i) |
| 58 | + seg[i].clear(); |
| 59 | + build(1, 1, n); |
| 60 | + |
| 61 | + int ret = 0; |
| 62 | + for(int i = 1; i <= n; ++i) { |
| 63 | + s = i, e = mx[a[i]], tar = a[i]; |
| 64 | + ret = max(ret, get(1, 1, n) + 1); |
| 65 | + } |
| 66 | + |
| 67 | + return ret; |
| 68 | +} |
| 69 | + |
| 70 | +int main() { |
| 71 | + scanf("%d", &n); |
| 72 | + for(int i = 1; i <= n; ++i) |
| 73 | + scanf("%d", a + i); |
| 74 | + |
| 75 | + if(n == 1) { |
| 76 | + puts("1"); |
| 77 | + return 0; |
| 78 | + } |
| 79 | + |
| 80 | + int r1 = solve(); |
| 81 | + reverse(a + 1, a + n + 1); |
| 82 | + int r2 = solve(); |
| 83 | + printf("%d\n", max(r1, r2)); |
| 84 | + |
| 85 | + return 0; |
| 86 | +} |
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