HDU 1159 Common Subsequence (动规+最长公共子序列)

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22698    Accepted Submission(s): 9967

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

 

Source

Southeastern Europe 2003

 

Recommend

Ignatius   |   We have carefully selected several similar problems for you:  1176 1058 1421 1160 1978
题目大意：求最长公共子序列。
在一些细节上有借鉴的地方，其它没什么了。

代码：

#include <iostream>
#include <string.h>
using namespace std;
#define M 1000
#define max(a,b) (a>b?a:b)
char ma1[M],ma2[M];
int dp[M][M];
int main(int i,int j,int k)
{ 
    int l1,l2;
  while(scanf("%s%s",ma1+1,ma2+1)!=EOF)         //这是个有意思的地方，由于在后面要用动规从1->l1，所以字符串从1開始。
  {   
      memset(dp,0,sizeof(dp));
      l1=strlen(ma1+1);                        //strlen是測字符串长度，没办法，还要+1.
      l2=strlen(ma2+1);

      for(i=1;i<=l1;i++)                       //这里就是模板。不说明了。

          for(j=1;j<=l2;j++)
          {
            if(ma1[i]==ma2[j]) dp[i][j]=dp[i-1][j-1]+1;
            else dp[i][j]=max(dp[i][j-1],dp[i-1][j]);
          }
   printf("%d
",dp[l1][l2]);
  }
  return 0;
}