uva 10635 - Prince and Princess
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 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 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 |
这道题很显然是LCS，但数据过大，因此O（n^2）的算法必然超时，其实LCS有一个不完全的O（nlogn）的算法。说它不完全，其实是有原因的，这个算法的复杂度是不定的，最坏情况O（n^2log(n^2))。但这题非常特殊，每个串中每个元素均只出现一次，因此复杂度就是严格O（nlogn）。
这个算法思想很简单，就是将b数组中的元素用一串递减下标代替，例如b[1]=5，那么查找5在a数组中出现的位置（递减顺序），比如a[5]=5,a[3]=5,a[2]=5,其余a[i]均不等于5.
那么b[1]用5,3,2代替，依次构造出一个新序列，对于a和b的LCS就等价于新序列的LIS，要知道LIS是有复杂度为O（nlogn）的算法的。至于为什么这样？可以证明任何一个递增子序列一一对应了a和b的一个公共子序列。因此最大长度是一样的。
代码：
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int a[62510],b[62510],dp[62510];
int main()
{
int t,m,n,x,cas=1;
scanf("%d",&t);
while(t--){
scanf("%*d%d%d",&m,&n);
memset(a,0,sizeof a);
for(int i=1;i<=m+1;i++){
scanf("%d",&x);
a[x]=i;
}
for(int i=1;i<=n+1;i++){
scanf("%d",&x);
b[i]=a[x];
}
int tot=0;
for(int i=1;i<=n+1;i++){
if(!b[i]) continue;
if(tot==0||b[i]>dp[tot-1]) dp[tot++]=b[i];
else{
int id=lower_bound(dp,dp+tot,b[i])-dp;
dp[id]=b[i];
}
}
printf("Case %d: %d\n",cas++,tot);
}
return 0;
}
注意：0实际表示不存在，对0要跳过。