【AHOI2006】基因匹配

题面

题解

众所周知，最长公共子序列的$dp$是$\text{O}(n^2)$，

但是每一个数字只重复$5$遍，那么我们暴力匹配$25n$个点对

那么我们就可以将其变成求最长上升子序列

用二分栈或者树状数组求解即可。

代码

#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#include<vector>
#define RG register

inline int read()
{
	int data = 0, w = 1; char ch = getchar();
	while(ch != '-' && (!isdigit(ch))) ch = getchar();
	if(ch == '-') w = -1, ch = getchar();
	while(isdigit(ch)) data = data * 10 + (ch ^ 48), ch = getchar();
	return data * w;
}

const int maxn(20010), maxm(500010);
int n, m, a[maxm], c[maxm], f[maxm], cnt, ans;
std::vector<int> G[maxn];

void add(int x, int v) { while(x <= m) c[x] = std::max(c[x], v), x += x & -x; }
int query(int x)
{
	int ans = 0;
	while(x) ans = std::max(ans, c[x]), x -= x & -x;
	return ans;
}

int main()
{
	m = (n = read()) * 5;
	for(RG int i = 1; i <= m; i++)
		G[read()].push_back(i);
	for(RG int i = 1, x; i <= m; i++)
	{
		x = read();
		for(RG int j = 4; ~j; j--) a[++cnt] = G[x][j];
	}
	for(RG int i = 1; i <= cnt; i++)
		f[i] = query(a[i] - 1) + 1, add(a[i], f[i]),
		ans = std::max(ans, f[i]);
	printf("%d\n", ans);
	return 0;
}