Codeforces 490F Treeland Tour(离散化 + 线段树合并)
题目链接 Treeland Tour
题目就是让你求树上LIS
先离散化,然后再线段树上操作。一些细节需要注意一下。
#include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i(a); i <= (b); ++i) #define dec(i, a, b) for (int i(a); i >= (b); --i) typedef long long LL; const int N = 200010; int root[N]; int ls[N * 40], rs[N * 40], lis[N * 40], lds[N * 40]; int ncnt, n, ans; int ret = 0; vector <int> v[N]; int val[N]; int sx[N]; int icnt = 0; void merge(int &x, int y){ if (!x || !y){ x = x + y; return; } lis[x] = max(lis[x], lis[y]); lds[x] = max(lds[x], lds[y]); ret = max(ret, max(lis[ls[x]] + lds[rs[y]], lds[rs[x]] + lis[ls[y]])); merge(ls[x], ls[y]); merge(rs[x], rs[y]); } void modify(int &x, int l, int r, int t, int v, int *a){ if (!x) x = ++ncnt; a[x] = max(a[x], v); if (l == r) return; int mid = (l + r) >> 1; if (t <= mid) modify(ls[x], l, mid, t, v, a); else modify(rs[x], mid + 1, r, t, v, a); } int query(int x, int l, int r, int ql, int qr, int *a){ if (l > r) return 0; if (!x) return 0; if (ql <= l && r <= qr) return a[x]; int ret = 0, mid = (l + r) >> 1; if (ql <= mid) ret = max(ret, query(ls[x], l, mid, ql, qr, a)); if (qr > mid) ret = max(ret, query(rs[x], mid + 1, r, ql, qr, a)); return ret; } void dfs(int x, int fa){ for (auto u : v[x]){ if (u == fa) continue; dfs(u, x); } ret = 0; int nlis = 0, nlds = 0, ilis, ilds; for (auto u : v[x]){ if (u == fa) continue; ilis = query(root[u], 1, icnt, 1, val[x] - 1, lis); ilds = query(root[u], 1, icnt, val[x] + 1, icnt, lds); merge(root[x], root[u]); ans = max(ans, ilis + nlds + 1); ans = max(ans, ilds + nlis + 1); nlis = max(nlis, ilis); nlds = max(nlds, ilds); } ans = max(ans, ret); modify(root[x], 1, icnt, val[x], nlis + 1, lis); modify(root[x], 1, icnt, val[x], nlds + 1, lds); } int main(){ scanf("%d", &n); rep(i, 1, n){ scanf("%d", val + i); sx[++icnt] = val[i]; } sort(sx + 1, sx + icnt + 1); icnt = unique(sx + 1, sx + icnt + 1) - sx - 1; rep(i, 1, n) val[i] = lower_bound(sx + 1, sx + icnt + 1, val[i]) - sx; rep(i, 2, n){ int x, y; scanf("%d%d", &x, &y); v[x].push_back(y); v[y].push_back(x); } dfs(1, 0); printf("%d\n", ans); return 0; }