经典DP水题G

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, x _ij = 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         abfcabprogramming    contest abcd           mnp

Sample Output

420

题意：从两个字符串中找相同的上升最大子串，输出长度。

思路：dp[i][j]，两层循环，表示当时的最大长度。

如果遍历到相等时，if (a[i - 1] == b[j - 1])     dp[i][j] = dp[i - 1][j - 1] + 1;否则取其两边大的一个。

AC java 代码：

import java.util.Scanner;


public class sdupractice0724DPG {
//3276 KB 282 ms
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
while (scan.hasNext()) {
String aa = scan.next();
String bb = scan.next();
char[] a = aa.toCharArray();
char[] b = bb.toCharArray();
int a1 = a.length;
int b1 = b.length;
int[][] dp = new int[a1+5][b1+5];
for (int i = 1; i <= a1; i++) {
for (int j = 1; j <= b1; j++) {
if (a[i - 1] == b[j - 1])
dp[i][j] = dp[i - 1][j - 1] + 1;
else {
if (dp[i - 1][j] > dp[i][j - 1])
dp[i][j] = dp[i - 1][j];
else
dp[i][j] = dp[i][j - 1];
}
}
}
System.out.println(dp[a1][b1]);
}
}
}