poj 1458 Common Subsequence(dp)
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 46630 | Accepted: 19154 |
Description
Input
Output
Sample Input
abcfbc abfcab
programming contest
abcd mnp
Sample Output
4
2
0
Java AC 代码:
import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in);
String first = "";
String second = "";
while(!(first = sc.next()).equals("") && !(second = sc.next()).equals("")) {
char[] firstArray = first.toCharArray();
char[] secondArray = second.toCharArray();
int firstLen = first.length();
int secondLen = second.length();
int[][] subMaxLen = new int[firstLen + 1][secondLen + 1]; //这里设置成长度加1的,是为了防止下面 i-1 j-1的时候数组越界。
for(int i = 1; i <= firstLen; i++)
for(int j = 1; j <= secondLen; j++) {
if(firstArray[i - 1] == secondArray[j - 1])
subMaxLen[i][j] = subMaxLen[i - 1][j - 1] + 1;
else
subMaxLen[i][j] = (subMaxLen[i - 1][j] > subMaxLen[i][j - 1] ? subMaxLen[i - 1][j] : subMaxLen[i][j - 1]);
}
System.out.println(subMaxLen[firstLen][secondLen]);
} } }
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