zoukankan      html  css  js  c++  java
  • hdu 1159 Common Subsequence(最长公共子序列)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1159

    Common Subsequence

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 37551    Accepted Submission(s): 17206


    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
     
    题目大意:求两个串的最长公共子序列长度。
     
    解题思路:套用模板。
     
    AC代码:
     1 #include <stdio.h>
     2 #include <string.h>
     3 #include <algorithm>
     4 
     5 using namespace std;
     6 
     7 char x[1000], y[1000];
     8 int b[1000][1000], c[1000][1000];
     9 void LCS()
    10 {
    11     int i,j;
    12     int len1 = strlen(x);
    13     int len2 = strlen(y);
    14     for (i = 0; i <= len1; i ++) b[i][0] = 0;
    15     for (j = 0; j <= len2; j ++) b[0][j] = 0;
    16     
    17     for (i = 1; i <= len1; i ++)
    18     {
    19         for (j = 1; j <= len2; j ++)
    20         {
    21             if (x[i-1] == y[j-1])
    22                 b[i][j] = b[i-1][j-1] + 1;
    23             else
    24                 b[i][j] = max(b[i-1][j],b[i][j-1]);
    25         }
    26     }
    27     printf("%d
    ",b[len1][len2]);
    28 }
    29 
    30 int main ()
    31 {
    32     while (scanf("%s%s",x,y)!=EOF)
    33     {
    34         LCS();
    35     }
    36     return 0;
    37 }
  • 相关阅读:
    pat 1029. Median (25)
    pat 1040. Longest Symmetric String (25)
    pat 1037. Magic Coupon (25)
    pat 1058. A+B in Hogwarts (20)
    pat 1083. List Grades (25)
    pat 1054. The Dominant Color (20)
    pat 1042. Shuffling Machine (20)
    pat 1061. Dating (20)
    *分支-11. 计算工资
    分支-10. 计算个人所得税
  • 原文地址:https://www.cnblogs.com/yoke/p/6657079.html
Copyright © 2011-2022 走看看