HDU-1159 Common Subsequence 最长上升子序列

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9595    Accepted Submission(s): 3923

Problem Description

A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = <x1, x2, ..., xm> another sequence Z = <z1, z2, ..., zk> is a subsequence of X if there exists a strictly increasing sequence <i1, i2, ..., ik> of indices of X such that for all j = 1,2,...,k, xij = zj. For example, Z = <a, b, f, c> is a subsequence of X = <a, b, c, f, b, c> with index sequence <1, 2, 4, 6>. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y. 
The program input is from a text file. Each data set in the file contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct. For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line. 

 

 

Sample Input

abcfbc abfcab programming contest abcd mnp

 

 

Sample Output

4 2 0

 

 

 

　　前面写了一份代码，一直不知到哪里处理错了，后来参考了别人的代码后发现了一个别人都是那样写，但是我却一直没注意的地方，那就是动态的方程在第一个元素的相等的时，dp[0][0] = dp[-1][-1] + 1, 天哪，这肯定就会出错了。在处理时可以选择字符的读取从第一个位置开始，或者把 i 号字符的状态存储到i+1号位置去，这样就从1号开始处理了，判定是就是 s1[i-1] == s1[j-1] ?

　　代码如下：

 1 #include <cstring>

 2 #include <cstdlib>

 3 #include <cstdio>

 4 #define Max( a, b ) (a) > (b) ? (a) : (b)

 5 using namespace std;

 6 

 7 char s1[1005], s2[1005];

 8 

 9 int dp[1005][1005];

10 

11 int main()

12 {

13     int len1, len2;

14     while( scanf( "%s %s", s1, s2 ) != EOF )

15     {

16         memset( dp, 0, sizeof(dp) );

17         len1 = strlen( s1 ), len2 = strlen( s2 );

18         for( int i = 1; i <= len1; ++i )

19         {

20             for( int j = 1; j <= len2; ++j )

21             {

22                 if( s1[i-1] == s2[j-1] )

23                 {

24                     dp[i][j] = dp[i-1][j-1] + 1;

25                 }

26                 else

27                 {

28                     dp[i][j] = Max ( dp[i-1][j], dp[i][j-1] );

29                 }

30             }

31         }

32         printf( "%d
", dp[len1][len2] );

33     }

34     return 0;

35 }

　　第二种处理方法：

#include <cstring>
#include <cstdlib>
#include <cstdio>
#define Max( a, b ) (a) > (b) ? (a) : (b)
using namespace std;

char s1[1005], s2[1005];

int dp[1005][1005];

int main()
{
    int len1, len2;
    while( scanf( "%s %s", s1+1, s2+1 ) != EOF )
    {
        memset( dp, 0, sizeof(dp) );
        len1 = strlen( s1+1 ), len2 = strlen( s2+1 );
        for( int i = 1; i <= len1; ++i )
        {
            for( int j = 1; j <= len2; ++j )
            {
                if( s1[i] == s2[j] )
                {
                    dp[i][j] = dp[i-1][j-1] + 1;
                }
                else
                {
                    dp[i][j] = Max ( dp[i-1][j], dp[i][j-1] );
                }
            }
        }
        printf( "%d ", dp[len1][len2] );
    }