CF568E Longest Increasing Subsequence 题解

CF568E Longest Increasing Subsequence

首先重复填数对答案没有影响\(,\)所以先忽略这个条件\(.\)

记\(f_i\)为 以i为结尾的LIS的长度\(,\) \(g_i\) 为以i为结尾的LIS的上一位\(.\)

记\(Min_x\)为 当前长度为\(x\)的LIS的最小的结尾 \(pos_x\)当前长度为\(x\)的LIS的最小的结尾的位置\(.\)

这些都可以简单\(dp\)出来\(.\)

然后我们考虑怎么构造方案\(.\)

不是\(-1\)的位置\(x\)我们可以直接跳到\(g_x\) \(.\)

如果\(x\)所在的位置是\(-1,\)那么我们优先找\(a_y ≠ -1 ,\)且\(a_y<a_x,f_y=\)当前的长度\(-1,\)那么我们就跳到位置\(y.\)

否则\(,\)我们就只能跳到最近的一个\(-1\)的位置\(.\)

\(O(n\log n+m\log m+(n+m)k)\)

代码\(:\)

#include <bits/stdc++.h>
using namespace std;
template <typename T> void read(T &x){
	x = 0; int f = 1; char ch = getchar();
	while (!isdigit(ch)) {if (ch == '-') f = -1; ch = getchar();}
	while (isdigit(ch)) {x = x * 10 + ch - '0'; ch = getchar();}
	x *= f;
}
inline void write(int x){if (x > 9) write(x/10); putchar(x%10+'0'); }
const int N = 100005,M = 100005,INF = 1e9 + 9;
int mn[N],pos[N],mx;
int n,a[N],f[N],g[N],m,b[M];
int main(){
	register int i,j,p;
	read(n); for (i = 1; i <= n; ++i) read(a[i]),mn[i] = INF; a[++n] = INF-1; mn[n] = INF;
	read(m); for (i = 1; i <= m; ++i) read(b[i]); sort(b+1,b+m+1);
	for (i = 1; i <= n; ++i){
		if (a[i] != -1){
			p = lower_bound(mn,mn+mx+1,a[i]) - mn; if (p > mx) ++mx;
			if (a[i] < mn[p]) mn[p] = a[i],pos[p] = i;
			f[i] = p,g[i] = pos[p-1]; continue;
		}
		for (p = mx+1,j = m; j ; --j){
			while (p && b[j] <= mn[p-1]) --p;
			if (b[j] < mn[p]) mn[p] = b[j],pos[p] = i;
		}
		f[i] = -1; if (pos[mx+1]) ++mx;
	}
	int now = n,nowlen = f[n],nowv = a[n];
	while (nowlen){
		if (a[now] != -1){ --nowlen,nowv = a[now],now = g[now]; continue; }
		for (i = m; i ; --i) if (b[i] != -1 && b[i] < nowv){ a[now] = b[i]; b[i] = -1; break; }
		for (i = now-1; i ; --i) if (f[i] == nowlen-1 && a[i] < a[now]){ g[now] = i; break; }
		if (!g[now]) for (i = now-1; i ; --i) if (a[i] == -1){ g[now] = i; break; }
		--nowlen,nowv = a[now],now = g[now];
	}
	for (i = 1; i <= n; ++i) if (a[i] == -1)
	for (j = 1; j <= m; ++j) if (b[j] != -1){ a[i] = b[j]; b[j] = -1; break; }
	for (i = 1; i < n; ++i) write(a[i]),putchar(i<n?' ':'\n');
	return 0;
}