题目链接

BZOJ5158

题解

题中所给的最长上升子序列其实就是一个限制条件

我们要构造出最大的以\(i\)开头的最长下降子序列，就需要编号大的点的权值尽量小

相同时当然就没有贡献，所以我们不妨令权值为一个\(1\)到\(n\)的排列

考虑如何满足限制条件

对于所有\(a[i] = v\)的点，点与点之间一定是单调下降的，连边

对于位置\(i\)，如果\(a[i] = v\)，那么\(i\)至少要比上一个\(a[j] = v - 1\)的位置大，连边

然后用大顶堆拓扑排序即可得到每个点最优的权值

\(O(nlogn)\)求一遍最长下降子序列长度即可

#include<iostream>

#include<cstdio>

#include<cmath>

#include<queue>

#include<cstring>

#include<algorithm>

#define LL long long int

#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)

#define REP(i,n) for (int i = 1; i <= (n); i++)

#define cls(s) memset(s,0x3f3f3f3f,sizeof(s))

using namespace std;

const int maxn = 100005,maxm = 100005,INF = 1000000000;

inline int read(){

	int out = 0,flag = 1; char c = getchar();

	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}

	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}

	return out * flag;

}

int h[maxn],ne = 2,de[maxn];

struct EDGE{int to,nxt;}ed[maxn << 1];

inline void build(int u,int v){

	ed[ne] = (EDGE){v,h[u]}; h[u] = ne++;

	de[v]++;

}

int last[maxn],n,a[maxn],cnt,x[maxn],f[maxn],bac[maxn];

priority_queue<int> q;

void work(){

	REP(i,n) if (!de[i]) q.push(i);

	int u;

	while (!q.empty()){

		u = q.top(); q.pop();

		x[u] = ++cnt;

		Redge(u) if (!(--de[to = ed[k].to])) q.push(to);

	}

}

int getn(int t){

	int l = 0,r = n,mid;

	while (l < r){

		mid = (l + r + 1) >> 1;

		if (bac[mid] < t) l = mid;

		else r = mid - 1;

	}

	return l;

}

void cal(){

	cls(bac); bac[0] = 0;

	for (int i = n; i; i--){

		f[i] = getn(x[i]) + 1;

		bac[f[i]] = min(bac[f[i]],x[i]);

	}

	LL ans = 0;

	REP(i,n) ans += f[i];

	printf("%lld\n",ans);

}

int main(){

	n = read();

	REP(i,n) a[i] = read();

	for (int i = 1; i <= n; i++){

		if (a[i] > 1) build(last[a[i] - 1],i);

		if (last[a[i]]) build(i,last[a[i]]);

		last[a[i]] = i;

	}

	work();

	cal();

	return 0;

}