zoukankan      html  css  js  c++  java
  • Common Subsequence

    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.
    Input
    abcfbc abfcab programming contest abcd mnp
    Output
    4 2 0
    SampleInput
     1 #include<stdio.h>
     2 #include<string.h>
     3 #define max 505
     4 int main()
     5 {
     6     char s1[max],s2[max];
     7     int matrix[max][max],len1,len2;
     8     int i,j;
     9     while(scanf("%s%s",s1,s2)!=EOF)
    10     {
    11         len1=strlen(s1);
    12         len2=strlen(s2);
    13         for(i=0; i<=len1; i++)matrix[i][0]=0;
    14         for(j=0; j<=len2; j++)matrix[0][j]=0;
    15         for(i=1; i<=len1; i++)
    16         {
    17             for(j=1; j<=len2; j++)
    18             {
    19                 if(s1[i-1]==s2[j-1])
    20                 {
    21                     matrix[i][j]=matrix[i-1][j-1]+1;
    22                 }
    23                 else
    24                 {
    25                     if(matrix[i-1][j]>=matrix[i][j-1])
    26                     {
    27                         matrix[i][j]=matrix[i-1][j];
    28                     }
    29                     else
    30                     {
    31                         matrix[i][j]=matrix[i][j-1];
    32                     }
    33                 }
    34             }
    35         }
    36         printf("%d
    ",matrix[len1][len2]);
    37     }
    38     return 0;
    39 }
    View Code
    abcfbc abfcab
    programming contest 
    abcd mnp
    SampleOutput
    4
    2
    0
    题目大意:给出两个字符串,求两个字符串的最长公共字串。
  • 相关阅读:
    LruCache 原理
    线程间通信, 进程间通信
    安卓 权限 规则
    android 捕获所有异常 未捕获的异常
    serializable parcelable
    android intent 传递 二进制数据
    apk安装 卸载 原理
    ARGB 8888 内存大小
    dalvik 基于 jvm 的改进
    查看 MySQL 数据库中每个表占用的空间大小
  • 原文地址:https://www.cnblogs.com/to-creat/p/4891203.html
Copyright © 2011-2022 走看看