poj1458公共子序列 C语言

/*Common Subsequence
Time Limit: 1000MS

Memory Limit: 10000K
Total Submissions: 56416

Accepted: 23516
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.
Input
The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.
Output
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*/
题意：求两个字符串的最大公共子序列。
公共子序列：顺序相同的序列，不一定是紧挨着，但是顺序一定相同。
假定我们有如下两个序列
S1: 1 2 3 4 5 6
S2: 4 5 6 7 8 9
S1和S2有一个最长公共子序列为 4 5 6一个子序列不一定必须是连续的，即中间可以被其他字符分开，但它们的顺序必须正确的。
最长公共子序列不一定只有一个。
S1: h e l l oS2: l e o nS1和S2有一个最长公共子序列为 eo
源程序如下：

 #include<stdio.h>
 #include<string.h>
 #define maxn 1005
 int n,m;//定义s1与s2的实际长度
 int dp[maxn][maxn];//把当前子序列的长度存入
 char s1[maxn],s2[maxn];//两个字符串
 int main()
 {int i,j;
     while(scanf("%s%s",s1,s2)==)
     {
         n=strlen(s1);//s1串长度
         m=strlen(s2);//s2串长度
         for( i=;i<=n;i++)//初始化dp
             for( j=;j<=m;j++)
                 dp[i][j]=;
     for( i=;i<=n;i++)//dp数组从1，1开始利用
         for( j=;j<=m;j++)
         {
             if(s1[i-]==s2[j-])/*如果s1的第i-1个字符与s2的第j-1个字符相同，
    那么s1[i-1]与s[j-1]的最长子序列是s[i-2]与s[j-2]的最长子序列加一*/
                 dp[i][j]=dp[i-][j-]+;
             else/*如果s1的第i-1个字符与s2的第j-1个字符不相同，
    那么s1[i-1]与s[j-1]的最长子序列是s[i-1]与s[j-2]的最长子序列
    或s[i-2]与s[j-1]的最长子序列之中最大的*/
                 if(dp[i-][j]>=dp[i][j-])
                  dp[i][j]=dp[i-][j];
                  else
                  dp[i][j]=dp[i][j-];
         }//循环递归，最后的dp[n][m]一定是s1[n],s2[m]最大的子序列长度
         printf("%d\n",dp[n][m]); 
     }
     return ;
 }

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