• Common Subsequence


    Common Subsequence
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 43153   Accepted: 17485

    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

    Source

    这么明显的一道动态规划,学长还讲过,这么low的题。。是不是傻
     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 
     5 using namespace std;
     6 
     7 #define N 500
     8 
     9 int main()
    10 {
    11     char str1[N], str2[N];
    12     int dp[N][N];
    13 
    14     while(~scanf("%s %s", str1, str2))
    15     {
    16         memset(dp, 0, sizeof(dp));
    17 
    18         int len1 = strlen(str1), len2 = strlen(str2);
    19 
    20         for(int i = 1; i <= len1; i++)
    21         {
    22             for(int j = 1; j <= len2; j++)
    23             {
    24                 if(str1[i-1] == str2[j-1])
    25                     dp[i][j] = dp[i-1][j-1]+1;
    26                 else
    27                     dp[i][j] = max(dp[i][j-1], dp[i-1][j]);
    28             }
    29         }
    30         printf("%d
    ", dp[len1][len2]);
    31     }
    32     return 0;
    33 }
    让未来到来 让过去过去
  • 相关阅读:
    前端使用crypto.js进行加密
    浅谈 Angular 项目实战
    YAML快速入门
    Preloading Your ASP.NET Applications
    ETL利器Kettle实战应用解析系列一【Kettle使用介绍】
    HBase
    hdfs知识点《转》
    Flume概念与原理、与Kafka优势对比《转》
    scrapy 快速入门
    比较好的算法网站
  • 原文地址:https://www.cnblogs.com/Tinamei/p/4717280.html
Copyright © 2020-2023  润新知