2016huasacm暑假集训训练五 C-Common Subsequence

题目链接： https://vjudge.net/contest/126708#problem/C

题意：这是一道求字符串的公共子串的最大长度的题目，用dp动态方程即可

if(a[i]==b[j]) dp[i][j]=dp[i-1][j-1]+1;

else dp[i][j]=(dp[i][j-1]>p[j][i-1]?dp[i][j-1]:p[j][i-1]);

这题主要是要注意两个字符串是同一行输入，题目没说字符串的长度，用c不是dp数组开大了编译不了就是开小了运行错误，错了几次就采用java的字符串来写，因为Java的数组可以根据字符串的长度来开，就不需要担心数组越界或者数组开小了；

AC代码：

 import java.io.BufferedInputStream;
 import java.util.Scanner;

 public class Main {
     static int dp[][];

     public static void main(String[] args) {

         Scanner s = new Scanner(new BufferedInputStream(System.in));
         while (s.hasNext()) {
             String a = s.next(), b = s.next();
             int len1 = a.length(), len2 = b.length();

             dp = new int[len1 + ][len2 + ];      //按字符串的长度开DP数组
             for (int i = ; i <= len1; i++)
                 for (int j = ; j <= len2; j++)
                     dp[i][j] = ;

             for (int i = ; i <= len1; i++)
                 for (int j = ; j <= len2; j++)
                     if (a.charAt(i - ) == b.charAt(j - )) {
                         dp[i][j] = dp[i - ][j - ] + ;
                     } else {
                         dp[i][j] = Math.max(dp[i - ][j], dp[i][j - ]);
                     }
             System.out.println(dp[len1][len2]);
         }
         s.close();
     }

 }