UVA 10635 Prince and Princess

大意:求出A串和B串的最长公共子序列。

思路:由于数据量比较大,p,q->250^2 = 62500.所以一般的解法是过不了的,在一看题目,发现两两元素各不相同,于是可以把A中元素重新编号为1~p+1。然后去B中找最长上升子序列即可,具体的可以手推一遍。

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <algorithm>
using namespace std;

const int MAXN = 260*260;
const int INF = 0x3f3f3f3f;

int A[MAXN], num[MAXN], S[MAXN];
int d[MAXN];

int n, p, q;

void init()
{
	n = 0;
	memset(num, 0, sizeof(num));
	memset(S, INF, sizeof(S));
	S[0] = 0;
}

void read_case()
{
	init();
	int x;
	scanf("%d%d%d", &x, &p, &q);
	for(int i = 1; i <= p+1; i++)
	{
		scanf("%d", &x);
		num[x] = i;
	}
	for(int i = 1; i <= q+1; i++)
	{
		scanf("%d", &x);
		if(num[x]) A[++n] = num[x];
	}
}

void dp()
{
	int ans = 0;
	for(int i = 1; i <= n; i++)
	{
		int pos = upper_bound(S, S+i, A[i]) - S;
		d[i] = pos;
		S[d[i]] = min(S[d[i]], A[i]);
		ans = max(ans, d[i]);
	}
	printf("%d\n", ans);
}

void solve()
{
	read_case();
	dp();
}

int main()
{
	int T, times = 0;
	scanf("%d", &T);
	while(T--)
	{
		printf("Case %d: ", ++times);
		solve();
	}
	return 0;
}


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