Prince and Princess - UVa 10635 dp-优快云博客

本文介绍了一种解决最长公共子序列问题的有效算法，通过将问题转化为最长递增子序列问题来减少计算复杂度，适用于大数据量的情况。

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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 x_p to denote the p-th square he enters, then x₁, x₂, ... x_p+1 are all different. Note that x₁ = 1 and x_p+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, ... y_q+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

3 6 7

1 7 5 4 8 3 9

1 4 3 5 6 2 8 9

Case 1: 4

题意：求最长公共子序列。

思路：因为数据量大，不能用n方的算法，要转化成最长递增子序列，方法是得到第二行的某一个数在第一行是第几个出现的，没有出现的可以忽略。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int num1[100010],num2[100010],dp[100010],len;
int solve(int k)
{ int l=1,r=len,mi;
  while(l<r)
  { mi=(l+r)/2;
    if(dp[mi]<k)
     l=mi+1;
    else
     r=mi;
  }
  return l;
}
int main()
{ int T,t,n,p,q,i,j,k;
  scanf("%d",&T);
  for(t=1;t<=T;t++)
  { scanf("%d%d%d",&n,&p,&q);
    memset(num1,0,sizeof(num1));
    p++;q++;n=0;len=0;
    for(i=1;i<=p;i++)
    { scanf("%d",&k);
      num1[k]=i;
    }
    for(i=1;i<=q;i++)
    { scanf("%d",&k);
      if(num1[k]>0)
       num2[++n]=num1[k];
    }
    for(i=1;i<=n;i++)
    { if(num2[i]>dp[len])
       dp[++len]=num2[i];
      else
       dp[solve(num2[i])]=num2[i];
    }
    printf("Case %d: %d\n",t,len);
  }
}