Uva 10635 简单dp

链接：戳这里

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

题意：T组数据,给出n,p,q (1<=p,q<=n*n)     p,q序列只包含(1~n*n)的数字，且唯一，问最长的公共子序列的长度

思路：首先我们分析一下样例

     3 6 7

P: 1 7 5 4 8 3 9

Q: 1 4 3 5 6 2 8 9

anw：(1 5 8 9) || (1,4,8,9)

一开始肯定是想暴力O(p*q)的思路,但是显然会超时啊

后面发现对p重新(1~p)编号{1,2,3,4,5,6},然后q去找p中对应的位置也就是{1,4,6,0,0,5,7}

应该可以看出来了吧 直接求LIS吧

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
int n,p,q,x,T;
int a[1000100],vis[1000100],anw[1000100];
int main(){
    scanf("%d",&T);
    for(int cas=1;cas<=T;cas++){
        mst(vis,0);
        scanf("%d%d%d",&n,&p,&q);
        p++;q++;n*=n;
        for(int i=1;i<=p;i++) {
            scanf("%d",&x);
            vis[x]=i;
        }
        for(int i=1;i<=q;i++) {
            scanf("%d",&x);
            a[i]=vis[x];
        }
        int tot=1;
        anw[0]=-1;
        for(int i=1;i<=q;i++){
            if(a[i]==0) continue;
            int pos=lower_bound(anw,anw+tot,a[i])-anw;
            if(pos>=tot) {
                anw[tot++]=a[i];
            } else {
                anw[pos]=a[i];
            }
        }
        printf("Case %d: %d\n",cas,tot-1);
    }
    return 0;
}