UVA10635 LCS 问题的 n*logn 解法

http://vjudge.net/contest/view.action?cid=53516#problem/B

Description

 

 

Problem E
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-thsquare 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

1 3 6 7 1 7 5 4 8 3 9 1 4 3 5 6 2 8 9

Case 1: 4

	Problemsetter: Man Rujia Liu Man, Member of Elite Problemsetters' Panel 

	Pictures drawn by Shahriar Manzoor, Member of Elite Problemsetters' Panel

题目的大意：

                      有两个长度分别为p+1和q+1的序列，每个序列中的各个元素都各不相同（这一点是保证本题有解的关键），且都是1~n*n之间的整数。两个序列的第一个元素均为1。求出A和B的最长公共子序列的长度。

解题思路：

               本题是LCS问题，但因为p和q高达250*250=62500，O（pq）的算法复杂度显得太慢。注意到A序列中所有的元素均不相同，因此可以把A中的元素重新编号为1~p+1

例如样例中A={1,7,5,4,8,3,9}，B={1,4,3,5,6,2,8,9}，因此把A重新编号为{1.2.3.4.5.6.7}，则B就是{1.4.6.3.0.0.5.7 }，其中0表示A中没有出现过（事实上可以直接删除这些元素，因为他们肯定不在LCS中）。这样，新的A和B的LCS实际上就是新的B的LIS。由于LIS可以在O（n*logn）的时间内解决，因此本题也可在O(nlogn)时间内得到解决。

代码如下：

 

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn =250*250;
const int inf=1000000000;
int s[maxn],g[maxn],d[maxn];
int num[maxn];//num[x]为整数x的新编号，num[x]=0表示x没有在A中出现过、
int N,p,q,x;
int main()
{
    int T,tt=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&N,&p,&q);
        memset(num,0,sizeof(num));
        for(int i=1;i<=p+1;i++)
        {
            scanf("%d",&x);
            num[x]=i;
        }
        int n=0;
        for(int i=0;i<q+1;i++)
        {
            scanf("%d",&x);
            if(num[x])
                s[n++]=num[x];
        }
        //求解s[0]...s[n-1]的LIS
        for(int i=1;i<=n;i++)
           g[i]=inf;
        int ans=0;
        for(int i=0;i<n;i++)
        {
            int k=lower_bound(g+1,g+n+1,s[i])-g;//在g[1]~g[n]中查找
            d[i]=k;
            g[k]=s[i];
            ans=max(ans,d[i]);
        }
        printf("Case %d: %d\n",++tt,ans);
    }
    return 0;
}