uva10635 LIS

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

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

题意:

有2个长度为p+1和q+1的序列，每个序列中的值都是唯一的，求2个序列的最长公共子序列。

 

思路:

如果做lcs复杂度为n^2，超时。我们队A数组内的元素重新编号，如果B数组中的元素之前在A中没有编号，那么说明这个元素不在A中出现，可以在B中删去

该元素。B元素中的值都有一个编号，并且该编号是根据A数组内元素的位置来给定的，所以现在只要求B数组的LIS。（B中有编号的A中一定出现，并且B中的值表示

这个数在A数组中的位置）

 

#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<string>
#include<vector>
#include<cstdio>
#include<time.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1000000000
#define MOD 1000000007
#define ll long long
#define lson l,m,rt<<1
#define key_value ch[ch[root][1]][0]
#define rson m+1,r,rt<<1|1
#define pi acos(-1.0)
using namespace std;
const int MAXN = 100010;
int num[MAXN],b[MAXN],n,q,p;
int g[MAXN];
int main(){
    int t,Case = 0;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%d",&n,&q,&p);
        memset(num,0,sizeof(num));
        for(int i = 1; i <= q+1; i++){
            int x;
            scanf("%d",&x);
            num[x] = i;
        }
        int cnt = 0;
        for(int i = 1; i <= p+1; i++){
            int x;
            scanf("%d",&x);
            if(num[x])b[cnt++] = num[x];
        }
        int ans = 0;
        g[0] = 0;
        for(int i = 1; i <= cnt; i++)g[i] = INF;
        for(int i = 0; i < cnt; i++){
            int tp = lower_bound(g+1,g+cnt+1,b[i]) - g;
            ans = max(ans,tp);
            g[tp] = min(g[tp],b[i]);
        }
        printf("Case %d: %d\n",++Case,ans);
    }
    return 0;
}