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POJ 1458 Common Subsequence(最长公共子序列)

时间:2019-06-01 19:56:26      阅读:93      评论:0      收藏:0      [点我收藏+]
Time Limit: 1000MS              Memory Limit: 10000K
Total Submissions: 67653        Accepted: 28245

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

解题思路:

重点在于状态转移方程的书写,这一题讲义PPT中画的图很好,言简意赅,我一开始想的是计算s2中以xk为终点的字串在s1中的公共子序列,但是发现自己对题意的理解有误,子串中的各字母是可以隔开的,因此逐个字符比较是最好的。既然是逐个字符相比较,那么自然也要考虑s1的位置。

AC代码:

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;

char s1[1000];
char s2[1000];
int maxLen[1000][1000];

int main()
{
    while (cin >> s1 >> s2)
    {
        int length1 = strlen(s1);
        int length2 = strlen(s2);
        for (int i = 0; i <= length1; i++)
            maxLen[i][0] = 0;
        for (int j = 0; j <= length1; j++)
            maxLen[0][j] = 0;
        for (int i = 1; i <= length1; i++)
        {
            for (int j = 1; j <= length2; j++)
            {
                if (s1[i - 1] == s2[j - 1])
                    maxLen[i][j] = maxLen[i - 1][j - 1] + 1;
                else
                    maxLen[i][j] = max(maxLen[i - 1][j], maxLen[i][j - 1]);
            }
        }
        cout << maxLen[length1][length2] << endl;
    }
    return 0;
}

 

POJ 1458 Common Subsequence(最长公共子序列)

原文:https://www.cnblogs.com/yun-an/p/10960726.html

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