POJ 1458 Common Subsequence_poj 1458解析-优快云博客

本文介绍了一个经典的动态规划问题——最长公共子序列(LCS)，通过给出详细的解题思路及示例代码，帮助读者理解如何利用二维动态规划数组解决此类问题。

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Common Subsequence

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_{i_j} = 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

题目大意：给出两个字符串，求最长公共子串LCS。

解题思路：非常经典的dp问题。定义dp[i][j]为串x的前i - 1位与串y的前j - 1位的最长公共子串，则对于x[i]和y[j]，若x[i] == y[j]，则dp[i + 1][j + 1] = dp[i][j] + 1；若x[i] != y[j]，则dp[i + 1][j + 1] = max(dp[i + 1][j],dp[i][j + 1])。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 300;

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

int main()
{
    int i,j,k;
    while(scanf("%s %s",x,y) != EOF){
        memset(dp,0,sizeof(0));
        int lx = strlen(x);
        int ly = strlen(y);
        for(i = 0;i < lx;i++){
            for(j = 0;j < ly;j++){
                if(x[i] == y[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]);
            }
        }
        printf("%d\n",dp[lx][ly]);
    }
    return 0;
}