zoukankan      html  css  js  c++  java
  • 最长公共子序列

                                             Common Subsequence

                                                    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                                             Total Submission(s): 9595    Accepted Submission(s): 3923


    Problem 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. 
    The program input is from a text file. Each data set in the file contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct. 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
     
     
     
    一个简单的动态规划的应用
    直接给代码
     1 #include <iostream>
     2 //#include <cstdio>
     3 #include <cstring>
     4 #define N 1005
     5 using namespace std;
     6 int dp[N][N];
     7 string s,ss;
     8 int main(){
     9     while(cin>>s>>ss){
    10         int n=s.length();
    11         int m=ss.length();
    12         memset(dp,0, sizeof(dp));
    13         for(int i=1;i<=n;i++){
    14             for(int j=1;j<=m;j++){
    15                 if(s[i-1]==ss[j-1])
    16                     dp[i][j]=dp[i-1][j-1]+1;
    17                 else{
    18                     dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
    19                 }
    20             }
    21         }
    22         cout<<dp[n][m]<<endl;
    23     }
    24     return 0;
    25 }
     
  • 相关阅读:
    Docker常用命令
    CentOS7.8源码安装nginx
    CentOS7.8搭建STF
    Mac关闭系统安全保护(提示无权删除/usr/bin目录下的文件时)
    Mac本地安装Tcloud-后端
    Mac后台运行进程-终端退出不影响其后台运行
    Docker快速部署TCloud云测试平台--前端
    Docker快速部署TCloud云测试平台--后端
    【项目】项目组件索引
    Netty 直接内存(堆外内存)溢出分析
  • 原文地址:https://www.cnblogs.com/zllwxm123/p/7232930.html
Copyright © 2011-2022 走看看