POJ - 1458 - Common Subsequence（dp & 公共子串）

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#dp & 最长公共子串

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本文详细解析了POJ-1458最长公共子序列问题，通过动态规划方法求解两个字符串的最长公共子序列长度。提供了完整的AC代码示例，展示了如何使用二维DP数组迭代计算最长公共子序列。

POJ - 1458 - Common Subsequence

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 abfcab
programming contest
abcd mnp
**Sample Output
4
2
0

题目链接
题意是让你从两个字符串中找最长的公共子串的长度。
这个我也是看的题解，貌似是一个模板题，需要一个dp二维数组，行列分别代表了题目中的两个字符串。dp[i][j]代表第一个字符串到第 i 个字符，第二个字符串到第 j 个字符的时候两个字符最长的公共子串的长度。所以，当s[i] == t[j]的时候，那么公共子串的长度又在原来的基础上大了1，否则就从dp[i + 1][j], dp[i][j + 1]前面找到的子串长度中选一个最长的。
则有了这样的代码

if(s[i] == t[j]) dp[i + 1][j + 1] = dp[i][j] + 1;
else dp[i + 1][j + 1] = max(dp[i + 1][j], dp[i][j + 1]);

AC代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
using namespace std;
const int maxn = 1e3 + 5;
int dp[maxn][maxn];

 int main()
 {
    int n;
    string s, t;
    while(cin >> s >> t)
    {
        memset(dp, 0, sizeof(dp));
        int ls = s.length();
        int lt = t.length();
        for(int i = 0; i < ls; i++)
            for(int j = 0; j < lt; j++)
            {
                if(s[i] == t[j])    dp[i + 1][j + 1] = dp[i][j] + 1;
                else dp[i + 1][j + 1] = max(dp[i + 1][j], dp[i][j + 1]);
            }
        cout << dp[ls][lt] << endl;
    }
    return 0;
}