zoukankan      html  css  js  c++  java
  • 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:
    基于STM32F103C8T6的超声波测距示例
    ST-LINK 到 SWD接线图
    MDK破解版下载
    CompletableFuture Quasar 等并发编程
    c++中参数传递和函数返回简析
    c++中冒号(:)和双冒号(::)的用法
    c++中,size_typt, size_t, ptrdiff_t 简介
    c++中,保证头文件只被编译一次,避免多重包含的方法
    时间序列分析之一次指数平滑法
    Openv2.1基本数据类型
  • 原文地址:https://www.cnblogs.com/Tinamei/p/4717280.html
Copyright © 2011-2022 走看看