poj1458 Common Subsequence

最新推荐文章于 2021-08-02 16:03:10 发布

原创最新推荐文章于 2021-08-02 16:03:10 发布 · 194 阅读

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本文介绍了一种解决最长公共子序列问题的算法，通过动态规划的方法，寻找两个字符串之间的最大长度公共子序列。该算法首先初始化一个二维数组用于存储中间结果，然后遍历两个字符串，比较字符并更新数组，最终输出最大公共子序列的长度。

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.

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

思路

a 串的第 i 个和 b 串的第 j 个相等，找 a 串前 i - 1 个和 b 串前 j - 1 个有多少个相等
a 串的第 i 个和 b 串的第 j 个不等：
- 找 a 串前 i 个和 b 串前 j - 1个有多少个相等
- 找 a 串前 i - 1 个和 b 串前 j 个有多少个相等

Code

#include <bits/stdc++.h>
#define N 1010
using namespace std;

int main()
{
    std::ios::sync_with_stdio(0);
    char t1[N] = {0},t2[N] = {0};
    while(scanf(" %s %s",t1,t2)!=EOF)
    {
        int i,j,dp[N][N];
        memset(dp,0,sizeof(dp));
        string s,t;
        s = " " + string(t1);
        t = " " + string(t2);
        for(i = 1; i < s.length(); i++)
        {
            for(j = 1; j < t.length(); j++)
            {
                if(s[i] == t[j])
                {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                }
                else
                {
                    dp[i][j] = max(dp[i - 1][j],dp[i][j - 1]);
                }
            }
        }
        cout<<dp[i - 1][j - 1]<<endl;
    }
    return 0;

}