HDU-1159 Common Subsequence(最长公共子序列)

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.

Input

abcfbc abfcab
programming contest
abcd mnp

Output

4
2
0

思路

大体的思路就是通过动态规划求两个序列的最长公共子序列，递推公式如下：

学习该算法的过程中主要参考了博客 动态规划基础篇之最长公共子序列问题

代码

#include
#include

int dp[1001][1001];
char x[1001], y[1001];

int main() {

    int len1, len2;
    while(~scanf("%s %s", x, y)) {
        len1 = strlen(x);
        len2 = strlen(y);
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= len1; i++) {
            for(int j = 1; j <= len2; j++) {
                if(x[i - 1] == y[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                }
                else {
                    dp[i][j] = dp[i - 1][j] > dp[i][j - 1] ? dp[i - 1][j] : dp[i][j - 1];
                }
            }
        }
        printf("%d\n", dp[len1][len2]);
    }
    return 0;
}