BZOJ 3173: [Tjoi2013]最长上升子序列 (线段树+BIT)

先用线段树预处理出每个数最终的位置.然后用BIT维护最长上升子序列就行了.

用线段树O(nlogn)O(nlogn)O(nlogn)预处理就直接倒着做,每次删去对应位置的数.具体看代码

CODE

#include<bits/stdc++.h>
using namespace std;
char cb[1<<15],*cs=cb,*ct=cb;
#define getc() (cs==ct&&(ct=(cs=cb)+fread(cb,1,1<<15,stdin),cs==ct)?0:*cs++)
template<class T>inline void read(T &res) {
	char ch; int flg = 1; while(!isdigit(ch=getc()))if(ch=='-')flg=-flg;
	for(res=ch-'0';isdigit(ch=getc());res=res*10+ch-'0'); res*=flg;
}
const int MAXN = 100005;
int n, m, x[MAXN], rk[MAXN], ans, sz[MAXN<<2], T[MAXN];
inline void chkmax(int &x, int y) { if(y > x) x = y; }
inline void upd(int x, int val) { for(; x <= n; x += x&-x) chkmax(T[x], val); }
inline int qsum(int x) { int re = 0; for(; x; x -= x&-x) chkmax(re, T[x]); return re; }
inline void upd(int i) { sz[i] = sz[i<<1] + sz[i<<1|1]; }
void build(int i, int l, int r) {
	if(l == r) { sz[i] = 1; return; }
	int mid = (l + r) >> 1;
	build(i<<1, l, mid);
	build(i<<1|1, mid+1, r);
	upd(i);
}
int query(int i, int l, int r, int k) {
	if(l == r) { sz[i] = 0; return l; }
	int mid = (l + r) >> 1, re;
	if(k <= sz[i<<1]) re = query(i<<1, l, mid, k);
	else re = query(i<<1|1, mid+1, r, k-sz[i<<1]);
	upd(i); return re;
}
int main() {
	read(n); build(1, 1, n);
	for(int i = 1; i <= n; ++i) read(x[i]), ++x[i];
	for(int i = n; i >= 1; --i) rk[i] = query(1, 1, n, x[i]); //!
	for(int i = 1, f; i <= n; ++i) {
		chkmax(ans, f = qsum(rk[i]) + 1);
		printf("%d\n", ans);
		upd(rk[i], f);
	}
}