P1108 低价购买

链接

https://www.luogu.com.cn/problem/P1108

思路

最长不降子序列。用dp方法是n²;非dp方法是nlog₂n。这里只能用dp方法,因为要统计有多少个。
具体思路就是用两个数组：f[N],c[N]。前者：正常记录最长子序列：f_i = max{f_j+1},i>j&&f[i]<f[j]。后者计数：1.对j<i && f[i] == f[j]+1,c[i]+=c[j]；2.对j<i &&f[i] == f[j],c[j]=0。第二条可以避免重复计数

同时注意不能马上加(参考错误代码,会加多)

WA代码

#include<bits/stdc++.h>
using namespace std;


#define IOS ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define int long long

using namespace std;


const int N = 5e3 + 10;

int n;
int a[N];
int f[N], c[N];
signed main()
{
	IOS;
	cin >> n;
	for (int i = 1; i <= n; i++) { cin >> a[i]; f[i]= 1; }
	int maxlen = 1;
	for (int i = 2; i <= n; i++)
	{
		for (int j = 1; j < i; j++)
		{
			if (a[j] > a[i])
			{
				f[i] = max(f[i], f[j] + 1);
			}
		}
		maxlen = max(maxlen, f[i]);
	}
	int ans = 0;
	c[1] = 1;
	for (int i = 2; i <= n; i++)
	{
		if (f[i] == 1)c[i] = 1;
		for (int j = 1; j < i; j++)
		{
			if (a[j] == a[i] and f[i] == f[j])
			{
				c[j] = 0;
			}
			else if (a[j] > a[i] and f[i] == f[j] + 1)
			{
				c[i] += c[j];
			}
		}//这里的区别,不能马上加,否则第二条原则就失效了
		if (f[i] == maxlen)ans += c[i];
	}
	cout << maxlen << ' ' << ans;
	return 0;
}

AC代码

#include<bits/stdc++.h>
using namespace std;


#define IOS ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define int long long

using namespace std;


const int N = 5e3 + 10;

int n;
int a[N];
int f[N], c[N];
signed main()
{
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n;
	for (int i = 1; i <= n; i++) { cin >> a[i]; f[i]= 1; }
	int maxlen = 1;
	int ans = 0;
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j < i; j++)
		{
			if (a[j] > a[i])
			{
				f[i] = max(f[i], f[j] + 1);
			}
		}
		maxlen = max(maxlen, f[i]);
		if (f[i] == 1)c[i] = 1;
		for (int j = 1; j < i; j++)
		{
			if (a[j] == a[i] && f[i] == f[j])
			{
				c[j] = 0;
			}
			else if (a[j] > a[i] and f[i] == f[j] + 1)
			{
				c[i] += c[j];
			}
		}
		
	}//这里的区别,不能马上加,否则第二条原则就失效了
	for (int i = 1; i <= n; i++)
	{
		if (f[i] == maxlen)
			ans += c[i];
	}
	cout << maxlen << ' ' << ans;
	return 0;
}