HDU 4681 String（2013多校8 1006题 DP）

String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 11    Accepted Submission(s): 4

Problem Description

Given 3 strings A, B, C, find the longest string D which satisfy the following rules:
a) D is the subsequence of A
b) D is the subsequence of B
c) C is the substring of D
Substring here means a consecutive subsequnce.
You need to output the length of D.

 

 

Input

The first line of the input contains an integer T(T = 20) which means the number of test cases.
For each test case, the first line only contains string A, the second line only contains string B, and the third only contains string C.
The length of each string will not exceed 1000, and string C should always be the subsequence of string A and string B.
All the letters in each string are in lowercase.

 

 

Output

For each test case, output Case #a: b. Here a means the number of case, and b means the length of D.

 

 

Sample Input

2 aaaaa aaaa aa abcdef acebdf cf

 

 

Sample Output

Case #1: 4 Case #2: 3

Hint

For test one, D is "aaaa", and for test two, D is "acf".

 

 

Source

2013 Multi-University Training Contest 8

 

 

Recommend

zhuyuanchen520

 

 

 

明显的DP题。

求最长公共子串，正和倒各求一次，得到dp和dp3

dp[i][j]表示str1的前i个字符 和 str2的前j个字符 最长的公共子串。

dp3[i][j]表示str1的后i个字符 和 str2的后j个字符 最长的公共子串。

 

dp1[i][j]表示str3的前j个字符，和str1的前i个字符匹配，最后的匹配起始位置，为-1表示不能匹配。

dp2一样

 

然后枚举str3在str1,str2匹配的终点

 

 

 1 /* ***********************************************
 2 Author        :kuangbin
 3 Created Time  :2013/8/15 12:34:55
 4 File Name     :F:\2013ACM练习\2013多校8\1006.cpp
 5 ************************************************ */
 6 
 7 #include <stdio.h>
 8 #include <string.h>
 9 #include <iostream>
10 #include <algorithm>
11 #include <vector>
12 #include <queue>
13 #include <set>
14 #include <map>
15 #include <string>
16 #include <math.h>
17 #include <stdlib.h>
18 #include <time.h>
19 using namespace std;
20 
21 const int MAXN = 1010;
22 char str1[MAXN],str2[MAXN],str3[MAXN];
23 int dp1[MAXN][MAXN];
24 int dp2[MAXN][MAXN];
25 int dp[MAXN][MAXN];
26 int dp3[MAXN][MAXN];
27 int main()
28 {
29     //freopen("in.txt","r",stdin);
30     //freopen("out.txt","w",stdout);
31     int T;
32     scanf("%d",&T);
33     int iCase = 0;
34     while(T--)
35     {
36         iCase++;
37         scanf("%s%s%s",str1,str2,str3);
38         int len1 = strlen(str1);
39         int len2 = strlen(str2);
40         int len3 = strlen(str3);
41         for(int i = 0; i <= len1;i++)
42             dp[i][0] = 0;
43         for(int i = 0;i <= len2;i++)
44             dp[0][i] = 0;
45         for(int i = 1;i <= len1;i++)
46             for(int j = 1;j <= len2;j++)
47             {
48                 dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
49                 if(str1[i-1] == str2[j-1])
50                     dp[i][j] = max(dp[i][j],dp[i-1][j-1]+1);
51             }
52         for(int i = 0; i <= len1;i++)
53             dp3[i][0] = 0;
54         for(int i = 0;i <= len2;i++)
55             dp3[0][i] = 0;
56         for(int i = 1;i <= len1;i++)
57             for(int j = 1;j <= len2;j++)
58             {
59                 dp3[i][j] = max(dp3[i-1][j],dp3[i][j-1]);
60                 if(str1[len1-i] == str2[len2-j])
61                     dp3[i][j] = max(dp3[i][j],dp3[i-1][j-1]+1);
62             }
63         for(int i = 1;i <= len3;i++)
64             dp1[0][i] = -1;
65         for(int i = 0;i <= len1;i++)
66             dp1[i][0] = i;
67         for(int i = 1;i <= len1;i++)
68             for(int j = 1;j <= len3;j++)
69             {
70                 if(str1[i-1] == str3[j-1])
71                     dp1[i][j] = dp1[i-1][j-1];
72                 else dp1[i][j] = dp1[i-1][j];
73             }
74         for(int i = 1;i <= len3;i++)
75             dp2[0][i] = -1;
76         for(int i = 0;i <= len2;i++)
77             dp2[i][0] = i;
78         for(int i = 1;i <= len2;i++)
79             for(int j = 1;j <= len3;j++)
80             {
81                 if(str2[i-1] == str3[j-1])
82                     dp2[i][j] = dp2[i-1][j-1];
83                 else dp2[i][j] = dp2[i-1][j];
84             }
85         int ans = 0;
86         for(int i = 0;i <= len1;i++)
87             for(int j = 0;j <= len2;j++)
88             {
89                 int t1 = dp1[len1-i][len3];
90                 int t2 = dp2[len2-j][len3];
91                 if(t1 == -1 || t2 == -1)continue;
92                 ans = max(ans,dp3[i][j]+dp[t1][t2]);
93             }
94         printf("Case #%d: %d\n",iCase,ans+len3);
95     }
96     return 0;
97 }