Problem D Prince and Princess Input: Standard Input
Output: Standard Output
Time Limit: 3 Seconds
 
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
제목: 왕자와 공주의 걸음걸이를 정한다.공동 주법을 구하다.
사고방식: 뚜렷한 LCS이지만 O(n^2) 알고리즘을 사용하면 n이 최대 250^2이기 때문에 시간을 초과할 수 있다.그래서 O(nlogn) 알고리즘을 사용합니다.아마 LIS 문제로 전환해서 LIS의 O(nlogn)로 구하고 있을 거예요.이 편을 자세히 보아라http://blog.csdn.net/accelerator_/article/details/11339459
그리고 이 문제에서 공주와 왕자는 반복되는 칸을 걷지 않기 때문이다.그래서 보관하기 편해요.
코드:

#include <stdio.h>
#include <string.h>
const int MAXN = 66666;
int t, n, na, nb, i, j, sb, a[MAXN], b[MAXN], save[MAXN], mid;

int two_set(int i, int j, int key) {
	while (i < j) {
		int mid = (i + j) / 2;
		if (save[mid] > key) j = mid;
		else i = mid + 1;
	}
	return j;
}
int main() {
	scanf("%d", &t);
	int tt = 1;
	while (t --) {
		memset(a, 0, sizeof(a));
		memset(b, 0, sizeof(b));
		memset(save, 0, sizeof(save));
		scanf("%d%d%d", &n, &na, &nb);
		na ++; nb ++; n = 1;
		for (i = 1; i <= na; i ++) {
			scanf("%d", &sb);
			a[sb] = i;
		}
		for (i = 1; i <= nb; i ++) {
			scanf("%d", &sb);
			if (a[sb]) {
				b[n ++] = a[sb];
			}
		}
		save[1] = b[1]; j = 1;
		for (i = 2; i < n; i ++) {
			if (b[i] > save[j]) {
				save[++ j] = b[i];
			}
			else {
				mid = two_set(1, j, b[i]);
				save[mid] = b[i];
			}
		}
		if (n == 1) j = 0;//           0.            
		printf("Case %d: %d
", tt ++, j);
	}
	return 0;
}