UVA 10653.Prince and Princess

题目

In an n * 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

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

题目大意

求一个最长公共子序列LCS，每个序列没有重复元素。

思路：

求解最长公共子序列。即，先扫一遍第一个序列，记录一下。然后扫一遍第二个序列，把与第一个序列重复的元素位置记录。最后求一个LIS。
这道题可以用map来对应，在这里我们用数组来做一个映射fa[a[i]]=i,就是代表其序列编号，然后将b中的元素b[i]对应的a[i]的序列编号输出到fb[]数组中，再求出fb[]的LIS即可。
这道题暗含序列不会重复。

代码

#include<cstdio>

#include<iostream>

#include<cstring>

#include<algorithm>

using namespace std;

#define cle(a) memset(a,0,sizeof(a))

const int Inf=0X3f3f3f3f,maxn=100000;

int n,a[maxn],b[maxn],dp[maxn],fa[maxn],fb[maxn];

int main(){

    int T; scanf("%d",&T);

    int all=T;

    while(T--){

        cle(a);cle(b);cle(fa);cle(fb);cle(dp);

        int n,p,q;

        scanf("%d%d%d",&n,&p,&q);

        for(int i=1;i<=p+1;i++){

            scanf("%d",&a[i]);

            fa[a[i]]=i;

        }

        for(int i=1;i<=q+1;i++){

            scanf("%d",&b[i]);

            fb[i]=fa[b[i]];

        }

        for(int i=1;i<=q+1;i++) dp[i]=1;

        for(int i=1;i<=q+1;i++){

            int max_=0;

            for(int j=1;j<i;j++)

                if(fb[j]<fb[i]) max_=max(max_,dp[j]);

            dp[i]+=max_;

        }

        int ans=0;

        for(int i=1;i<=q+1;i++)

            ans=max(ans,dp[i]);

        printf("Case %d: %d\n",all-T,ans);

    }

    return 0;

}