Uva10635 Prince and Princess

题目戳这里
这题如果用(f_{i,j})这样dp的话肯定过不了，必须另辟蹊径。题目说了数字不重复。我们先只留下两个数组共有的数字。然后我们处理出这样一个数组(S)，(S_i)表示(A_i)这个元素在(B)中的下标，然后模型转换就成为了求(S)中最长上升子序列了，这个(O(NlogN))的求法大家应该都会。这里我写的是树状数组版本的。

#include<iostream>
#include<cstdio>
#include<cstdlib>
using namespace std;

#define lowbit(x) (x&-x)
const int maxn = 250*250+10;
int pos[maxn],S[maxn],T,tree[maxn],N,P,Q,cnt,ans;

inline void ins(int a,int b) { for (;a <= N*N;a += lowbit(a)) tree[a] = max(tree[a],b); }
inline int calc(int a) { int ret = 0; for (;a;a -= lowbit(a)) ret = max(ret,tree[a]); return ret; }

inline int read()
{
	int ret = 0,f = 1; char ch;
	do ch = getchar(); while (!(ch >= '0'&&ch <= '9')&&ch != '-');
	if (ch == '-') ch = getchar(),f = -1;
	do ret = ret*10+ch-'0',ch = getchar(); while (ch >= '0'&&ch <= '9');
	return ret*f;
}

int main()
{
	freopen("10635.in","r",stdin);
	freopen("10635.out","w",stdout);
	scanf("%d",&T);
	for (int Case = 1;Case <= T;++Case)
	{
		printf("Case %d: ",Case);
		N = read(); P = read()+1; Q = read()+1; cnt = ans = 0;
		for (int i = 1;i <= N*N;++i) pos[i] = tree[i] = 0;
		for (int i = 1;i <= P;++i) pos[read()] = i;
		for (int i = 1,b;i <= Q;++i)
		{
			b = read();	
			if (pos[b]) S[++cnt] = pos[b];
		}
		for (int i = 1;i <= cnt;++i)
		{
			int f = calc(S[i]-1)+1;
			ans = max(ans,f); ins(S[i],f);
		}
		printf("%d
",ans);
	}
	fclose(stdin); fclose(stdout);
	return 0;
}