hdu 1159 Common Subsequence（LCS）

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37725    Accepted Submission(s):
17301

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

题目大意：

　　　　　输入两个字符串，输出这两个字符串的最长公共子序列长度。

解题思路：

　　　　　最长公共子序列模板题，算法详解：http://www.cnblogs.com/yoke/p/6686898.html

 #include <stdio.h>
 #include <string.h>

 char s1[],s2[];
 int x[][];   // 记录最长公共子序列
 int LCS()
 {
     int i,j;
     int l1 = strlen(s1);     // 计算字符串的长度
     int l2 = strlen(s2);
     memset(x,,sizeof(x));  // 初始化 过滤掉0的情况 

     for (i = ; i <= l1; i ++)
     {
         for (j = ; j <= l2; j ++)
         {
             if (s1[i-] == s2[j-])   // 相等的情况
                // 字符数组是从0开始的 所以这里要减 1
                 x[i][j] = x[i-][j-]+;
             else if(x[i-][j] >= x[i][j-])    // 不相等的时候选择 比较“左边”和“上边”选择较大的
                 x[i][j] = x[i-][j];
             else
                 x[i][j] = x[i][j-];
         }
     }
     return x[l1][l2];
 }
 int main ()
 {
     while (scanf("%s%s",s1,s2)!=EOF)
     {
         int len = LCS();
         printf("%d\n",len);
     }
     return ;
 }