uva 10635 Prince and Princess（DP）

uva 10635 Prince and Princess（DP）

In an n x n chessboard, Prince and Princess plays a game. The squares in the chessboard are numbered
1, 2, 3 ... n*n, as shown below:



Prince stands in square 1, make p jumps and finally reach square
n*n. He enters a square at most once. So if we use xp to denote the
p-th square he enters, then x1,
x2, ... xp+1 are all different. Note that
x1 = 1 and xp+1 = n*n. Princess does the similar thing - stands in square
1, make q jumps and finally reach square
n*n. We use y1, y2, ...
yq+1 to denote the sequence, and all
q+1 numbers are different.



Figure 2 belows show a 3x3 square, a possible route for Prince and a different route for Princess.



The Prince moves along the sequence: 1 --> 7 --> 5 --> 4 --> 8 --> 3 --> 9 (Black arrows), while the Princess moves along this sequence:
1 --> 4 --> 3 --> 5 --> 6 --> 2 --> 8 --> 9 (White arrow).

The King -- their father, has just come. "Why move separately? You are brother and sister!" said the King, "Ignore some jumps and make sure that you're always together."



For example, if the Prince ignores his 2nd, 3rd,
6th jump, he'll follow the route: 1 --> 4 --> 8 --> 9. If the Princess ignores her
3rd, 4th, 5th, 6th jump, she'll follow the same route:
1 --> 4 --> 8 --> 9, (The common route is shown in figure
3) thus satisfies the King, shown above. The King wants to know the longest route they can move together, could you tell him?



Input

The first line of the input contains a single integer
t(1 <= t <= 10), the number of test cases followed. For each case, the first line contains three integers
n, p, q(2 <= n <= 250, 1 <= p, q < n*n). The second line contains
p+1 different integers in the range [1..n*n], the sequence of the Prince. The third line contains
q+1 different integers in the range [1..n*n], the sequence of the Princess.

Output

For each test case, print the case number and the length of longest route. Look at the output for sample input for details.

Sample Input
Output for Sample Input

1 3 6 7 1 7 5 4 8 3 9 1 4 3 5 6 2 8 9

Case 1: 4

题目大意：两个数字序列，求两个序列的最长公共子序列。

解题思路：因为数据量比较大，所以要用
n*logn 的方法，生成B对应A的映射表，然后在映射表中找最长上升子序列，就是两个序列的最长公共子序列。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#define N 250 * 250 + 5 
using namespace std;
const int INF = 100000000;
int num
, s
, g
, d
;
int main() {
	int T, Case = 1;
	scanf("%d", &T);
	while (T--) {
		int n, p, q;
		scanf("%d %d %d", &n, &p, &q);	
		memset(num, 0, sizeof(num));
		int a, cnt = 0, ans = 0;
		for (int i = 1; i <= p + 1; i++) {
			scanf("%d", &a);
			num[a] = i;
		}
		for (int i = 0; i < q + 1; i++) {
			scanf("%d", &a);
			if (num[a]) {
				s[cnt++] = num[a]; //生成B对于A的映射表
			}
		}
		for (int i = 1; i <= cnt; i++) g[i] = INF;
		/*在B中查找最长上升子序列，就是A与B的最长公共子序列*/
		for (int i = 0; i < cnt; i++) {
			int temp = lower_bound(g + 1, g + cnt + 1, s[i]) - g;
			/*在g中找到第一个比s[i]大的数，替换它*/
			d[i] = temp;
			g[temp] = s[i];
			ans = max(ans, d[i]);
		}
		printf("Case %d: %d\n", Case++, ans);
	}
	return 0;
}