ZOJ 2432 Greatest Common Increasing Subsequence

LCIS+记录路径

LCIS：
dp[i][j] 表示数列a前i项，和数列b前j项且以b[ j ]结尾的LCIS。
当a[ i ]！= b[ j ]时，dp[ i ][ j ]==dp[i-1][ j ]
当a[ i ]==b[ j ]时，dp[ i ][ j ]==dp[i-1][1~j之间最大的] +1

最后统计 dp[len(a)][...]里最大的就是LICS

Greatest Common Increasing Subsequence

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : Accepted Submission(s) :

Special Judge

Problem Description

You are given two sequences of integer numbers. Write a program to determine their common increasing subsequence of maximal
possible length. Sequence S1, S2, ..., SN of length N is called an increasing subsequence of a sequence A1, A2, ..., AM of length M if there exist 1 <= i1 < i2 < ...< iN <= M such that Sj = Aij for all 1 <= j <= N, and Sj < Sj+1 for all 1 <= j < N.

Input

Each sequence is described with M - its length (1 <= M <= 500) and M integer numbers Ai (-2^31 <= Ai < 2^31) - the sequence itself.

Output

On the first line of the output print L - the length of the greatest common increasing subsequence of both sequences. On the second line print the subsequence itself. If there are several possible answers, output any of them.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Sample Input

1

5
1 4 2 5 -12
4
-12 1 2 4

Sample Output

2
1 4

 1 #include <iostream>
 2 #include <cstring>
 3 
 4 using namespace std;
 5 
 6 const int N=555;
 7 
 8 struct p
 9 {
10     int x,y;
11 }path[N][N];
12 
13 int a[N],b[N],dp[N][N];
14 
15 int main()
16 {
17     int t;
18     cin>>t;
19 while(t--)
20 {
21     int n,m;
22     cin>>n;
23     for(int i=1;i<=n;i++)
24         cin>>a[i];
25     cin>>m;
26     for(int j=1;j<=m;j++)
27         cin>>b[j];
28 
29     int maxn=0;
30     memset(dp,0,sizeof(dp));
31     for(int i=1;i<=n;i++)
32     {
33         maxn=0;
34         int tx=0,ty=0;
35         for(int j=1;j<=m;j++)
36         {
37             dp[i][j]=dp[i-1][j];
38             path[i][j].x=i-1;
39             path[i][j].y=j;
40             if(a[i]>b[j]&&maxn<dp[i-1][j])
41             {
42                 maxn=dp[i-1][j];
43                 tx=i-1;ty=j;
44             }
45             if(a[i]==b[j])
46             {
47                 dp[i][j]=maxn+1;
48                 path[i][j].x=tx;
49                 path[i][j].y=ty;
50             }
51         }
52     }
53 
54     maxn=-1;
55     int id;
56     for(int i=1;i<=m;i++)
57     {
58         if(maxn<dp[n][i])
59         {
60             maxn=dp[n][i];
61             id=i;
62         }
63     }
64 
65     int save[N];
66     int cnt=0;
67     int tx=n,ty=id;
68     while(dp[tx][ty]!=0)
69     {
70         int tmpx=path[tx][ty].x;
71         int tmpy=path[tx][ty].y;
72 
73         if(dp[tmpx][tmpy]!=dp[tx][ty])
74         {
75             save[cnt++]=b[ty];
76         }
77 
78         tx=tmpx;  ty=tmpy;
79     }
80 
81     cout<<maxn<<endl;
82     for(int i=cnt-1;i>=0;i--)
83     {
84         cout<<save[i]<<" ";
85     }
86     cout<<endl;
87 }
88 
89     return 0;
90 }