BZOJ 3173: [Tjoi2013]最长上升子序列( BST + LIS )
因为是从1~n插入的, 慢插入的对之前的没有影响, 所以我们可以用平衡树维护, 弄出最后的序列然后跑LIS就OK了 O(nlogn)
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#include<bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < n; ++i)
#define clr(x, c) memset(x, c, sizeof(x))
#define foreach(i, x) for(__typeof(x.begin()) i = x.begin(); i != x.end(); i++)
using namespace std;
const int maxn = 100009;
struct Node {
Node *ch[2], *p;
int s, v;
inline void upd() {
s = ch[0]->s + ch[1]->s + 1;
}
inline void setc(Node* t, int d) {
ch[d] = t;
t->p =this;
}
inline bool d() {
return this == p->ch[1];
}
} pool[maxn], *pt = pool, *root, *null;
Node* newNode(int _ = 0) {
pt->v = _; pt->s = 1;
pt->ch[0] = pt->ch[1] = pt->p = null;
return pt++;
}
void rot(Node* t) {
Node* p = t->p;
int d = t->d();
p->p->setc(t, p->d());
p->setc(t->ch[d ^ 1], d);
t->setc(p, d ^ 1);
p->upd();
if(p == root) root = t;
}
void splay(Node* t, Node* f = null) {
while(t->p != f) {
if(t->p->p != f)
t->d() != t->p->d() ? rot(t) : rot(t->p);
rot(t);
}
t->upd();
}
Node* select(int k) {
for(Node* t = root; ;) {
int s = t->ch[0]->s;
if(k == s) return t;
if(k > s)
k -= s + 1, t = t->ch[1];
else
t = t->ch[0];
}
}
void init() {
null = newNode();
null->ch[0] = null->ch[1] = null->p = null;
null->s = 0;
root = newNode(maxn);
root->setc(newNode(-maxn), 0);
root->upd();
}
int seq[maxn], N = 0, g[maxn], dp[maxn];
void dfs(Node* t) {
if(t == null) return;
dfs(t->ch[0]);
if(t->v != maxn && t->v != -maxn) seq[N++] = t->v;
dfs(t->ch[1]);
}
int main() {
freopen("test.in", "r", stdin);
init();
int n;
cin >> n;
rep(i, n) {
int v;
scanf("%d", &v);
Node *L = select(v), *R = select(v + 1);
splay(L); splay(R, L);
R->setc(newNode(i), 0);
R->upd(); L->upd();
g[i] = maxn;
}
dfs(root);
rep(i, n) {
int p = lower_bound(g, g + n, seq[i]) - g;
dp[seq[i]] = p + 1;
g[p] = seq[i];
}
int ans = 0;
rep(i, n) {
ans = max(ans, dp[i]);
printf("%d\n", ans);
}
return 0;
}
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3173: [Tjoi2013]最长上升子序列
Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 806 Solved: 432
[ Submit][ Status][ Discuss]
Description
给定一个序列，初始为空。现在我们将1到N的数字插入到序列中，每次将一个数字插入到一个特定的位置。每插入一个数字，我们都想知道此时最长上升子序列长度是多少？
Input
第一行一个整数N，表示我们要将1到N插入序列中，接下是N个数字，第k个数字Xk，表示我们将k插入到位置Xk（0<=Xk<=k-1,1<=k<=N）
Output
N行，第i行表示i插入Xi位置后序列的最长上升子序列的长度是多少。
Sample Input
3
0 0 2
0 0 2
Sample Output
1
1
2
1
2
HINT
100%的数据 n<=100000
Source