POJ 2127 Greatest Common Increasing Subsequence (最长公共上升子序列+记录路径)

Greatest Common Increasing Subsequence

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 8021		Accepted: 2138
Case Time Limit: 2000MS		Special Judge

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 (-231 <= Ai < 231 ) --- the sequence itself.
Output
On the first line of the output file 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.
Sample Input

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

Sample Output

2
1 4

思路：定义dp[i][j]表示第一个序列seq1[]到位置i，第二个序列seq2[]到位置j时的最长公共上升子序列长度。当seq1[i] != seq2[j]时，dp[i][j] = dp[i-1][j];否则更新dp[i][j]的值为之前最长的长度+1，同时用二维数组pre[i][j]存储上个最长长度是以seq2[]的哪个结尾的，假设之前的最长长度是dp[i'][j']，则pre[i][j] = j'。

View Code

1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #define SIZE 505
6
7 using namespace std;
8
9 int dp[SIZE][SIZE];
10 int pre[SIZE][SIZE];
11 int path[SIZE];
12 int seq1[SIZE],seq2[SIZE];
13 int N,M;
14
15 int main()
16 {
17     while(~scanf("%d",&N))
18     {
19         for(int i=1; i<=N; i++)
20             scanf("%d",&seq1[i]);
21         scanf("%d",&M);
22         for(int i=1; i<=M; i++)
23             scanf("%d",&seq2[i]);
24         int ans = -1;
25         int pos1,pos2;
26         memset(dp,0,sizeof(dp));
27         memset(pre,0,sizeof(pre));
28         memset(path,0,sizeof(path));
29         for(int i=1; i<=N; i++)
30         {
31             int maxi = 0,temp = 0;
32             for(int j=1; j<=M; j++)
33             {
34                 if(seq1[i] != seq2[j])
35                 {
36                     dp[i][j] = dp[i-1][j];
37                     if(seq1[i] > seq2[j] && maxi < dp[i][j])
38                     {
39                         maxi = dp[i][j];
40                         temp = j;
41                     }
42                 }
43                 if(seq1[i] == seq2[j])
44                 {
45                     dp[i][j] = maxi + 1;
46                     pre[i][j] = temp;
47                 }
48                 if(dp[i][j] > ans)
49                 {
50                     ans = dp[i][j];
51                     pos1 = i;
52                     pos2= j;
53                 }
54             }
55         }
56         printf("%d\n",ans);
57         int len = ans;
58         if(ans > 0)
59             path[ans--] = pos2;
60         while(ans && pos1 && pos2)
61         {
62             if(pre[pos1][pos2] > 0)
63             {
64                 path[ans--] = pre[pos1][pos2];
65                 pos2 = pre[pos1][pos2];
66             }
67             pos1--;
68         }
69         for(int i=1; i<len; i++)
70             printf("%d ",seq2[path[i]]);
71         printf("%d\n",seq2[path[len]]);
72     }
73     return 0;
74 }