HDU 1159 Common Subsequence

Common Subsequence

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

 

题目大意：求最长公共子序列

 

代码如下：

 1 #include <iostream>
 2 # include<cstdio>
 3 # include<cstring>
 4 using namespace std;
 5 int max(int a,int b)
 6 {
 7     return a>b ?a : b;
 8 }
 9 char s1[1005],s2[1005];
10 int dp[1005][1005];
11 int main()
12 {
13     int i,j,len1,len2;
14     while(scanf("%s%s",s1,s2)!=EOF)
15     {
16         len1 = strlen(s1);
17         len2 = strlen(s2);
18         int ans=0;
19         dp[0][0] = 0;
20         for(i=1; i<=len1; i++)
21             dp[i][0] = 0;
22         for(j=1; j<=len2; j++)
23             dp[0][j] = 0;
24         for(i=1; i<=len1; i++)
25         {
26             for(j=1; j<=len2; j++)
27             {
28                 if(s1[i-1]==s2[j-1])
29                     dp[i][j] = max(max(dp[i-1][j],dp[i][j-1]),dp[i-1][j-1]+1);
30                 else
31                     dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
32                 if(dp[i][j]>ans)
33                     ans= dp[i][j];
34             }
35         }
36         printf("%d\n",ans);
37     }
38     return 0;
39 }