UVA 10635 Prince and Princess—— 求LCS（最長公共子序列）轉換成 求LIS（最長遞增子序列）

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：

兩個沒有重複元素的數組a和b，求出a和b的最長公共子序列。

 

思路：

這題可以轉化爲最長遞增子序列來做。

建立一個數組 f1[ ] 作爲映射，將a序列映射到這個數組，映射方法爲 ： f1[ 序列元素 ] = 序列編號

如樣例就是：

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

然後，對這個映射將b序列映射出的新數組b2，求b2最長遞增子序列，即爲答案。

注意：只有沒有重複元素的序列，才能這麼做。

 

#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