Prince and Princess

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     

                      1

3 6 7

1 7 5 4 8 3 9

1 4 3 5 6 2 8 9

 

Output for Sample Input

Case 1: 4  

 

大意：

两个没有重复元素的数组a和b，求出a和b的最长公共子序列。

 

思路：

这题可以转化为最长递增子序列来做。

建立一个数组 f1[ ] 作为映射，将a序列映射到这个数组，映射方法为 ： f1[ 序列元素 ] = 序列编号

如样例就是：

f1[1] = 1; f[7] = 2; f[5] = 3; f[4] = 4; f[8] = 5; ...

然后，对这个映射将b序列映射出的新数组b2，求b2最长递增子序列，即为答案。

注意：只有没有重复元素的序列，才能这么做。

 

#include<cstdio> #include<cstdlib> #include<cmath> #include<cstring> #include<string> #include<iostream> #include<stack> #include<queue> #include<algorithm> #include<map> #include<list> #include<vector> #define INF 0x3f3f3f3f #define MAXN 252 using namespace std; int main() { int t, Case = 1; scanf("%d", &t); while(t--) { int i, j, n, an, bn, a[MAXN*MAXN] = {0}, b[MAXN*MAXN] = {0}; int b2[MAXN*MAXN] = {0}, f1[MAXN*MAXN] = {0}; scanf("%d%d%d", &n, &an, &bn); an++; bn++; for(i = 1; i <= an; i++) { scanf("%d", &a[i]); f1[a[i]] = i; } for(i = 1; i <= bn; i++) { scanf("%d", &b[i]); b2[i] = f1[b[i]]; } int dp[MAXN*MAXN]; for(i = 1; i <= bn; i++) { dp[i] = 1; } for(i = 1; i <= bn; i++) { int maxdp = 0; for(j = 1; j <= i-1; j++) { if(b2[j] < b2[i]) maxdp = max(maxdp, dp[j]); } dp[i] += maxdp; } int ans = 0; for(i = 1; i <= bn; i++) { ans = max(ans, dp[i]); } printf("Case %d: %d\n", Case++, ans); } return 0; } //1 //3 6 7 //1 7 5 4 8 3 9 //1 4 3 5 6 2 8 9