Bzoj1046: [HAOI2007]上升序列

题面

传送门

Sol

先求出最长上升序列,倒着求,然后贪心的往后选,选满足的
求最长上升序列我用的是树状数组

# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1e4 + 5);

IL ll Read(){
    RG char c = getchar(); RG ll x = 0, z = 1;
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int n, a[_], m, o[_], len, maxlen;
struct Data{
	int v, p;
	IL bool operator <(RG Data B) const{
		return v != B.v ? v < B.v : p > B.p;
	}
} bit[_], f[_];

IL void Add(RG int x, RG Data mx){
	for(; x <= len; x += x & -x) bit[x] = max(bit[x], mx);
}

IL Data Query(RG int x){
	RG Data ret = (Data){-1, 0};
	for(; x; x -= x & -x) ret = max(bit[x], ret);
	return ret;
}

int main(RG int argc, RG char* argv[]){
	n = Read();
	for(RG int i = 1; i <= n; ++i) o[++len] = a[i] = -Read();
	sort(o + 1, o + len + 1);
	len = unique(o + 1, o + len + 1) - o - 1;
	for(RG int i = 1; i <= n; ++i) a[i] = lower_bound(o + 1, o + len + 1, a[i]) - o;
	f[n] = (Data){1, 0}; Add(a[n], (Data){1, n});
	for(RG int i = n - 1; i; --i){
		f[i] = (Data){1, 0};
		RG Data mx = Query(a[i] - 1);
		f[i] = max(f[i], (Data){mx.v + 1, mx.p});
		Add(a[i], (Data){f[i].v, i});
		maxlen = max(maxlen, f[i].v);
	}
	m = Read();
	for(RG int i = 1; i <= m; ++i){
		RG int l = Read(), mx = -1e9;
		if(l > maxlen) puts("Impossible");
		else{
			for(RG int i = 1; i <= n && l; ++i)
				if(f[i].v >= l && -o[a[i]] > mx){
					--l;
					printf("%d", mx = -o[a[i]]);
					if(l != 0) putchar(' ');
				}
			puts("");
		}
	}
	return 0;
}

posted @ 2018-01-25 21:33  Cyhlnj  阅读(129)  评论(0编辑  收藏  举报