zoukankan      html  css  js  c++  java
  • POJ 1458 Common Subsequence

    Common Subsequence
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 33069   Accepted: 12966

    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
    解题方法:最长公共子序列,直接套用模版。
    #include <stdio.h>
    #include <string.h>
    #include <iostream>
    using namespace std;
    
    int main()
    {
        char str[300];
        char substr[300];
        int dp[300][300];
        while(scanf("%s %s", str, substr) != EOF)
        {
            int nLen = strlen(str);
            int nLen1 = strlen(substr);
            for (int i = 0; i <= nLen; i++)
            {
                dp[i][0] = 0;
            }
            for (int i = 0; i <= nLen1; i++)
            {
                dp[0][i] = 0;
            }
            for (int i = 1; i <= nLen; i++)
            {
                for (int j = 1; j <= nLen1; j++)
                {
                    if (str[i - 1] == substr[j - 1])
                    {
                        dp[i][j] = dp[i - 1][j - 1] + 1;
                    }
                    else
                    {
                        dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]);
                    }
                }
            }
            printf("%d
    ", dp[nLen][nLen1]);
        }
        return 0;
    }


  • 相关阅读:
    进军装饰器2/3
    进军装饰器1/3
    购物车
    多级菜单(高效版)
    工资管理系统
    多级菜单(低效版)
    用户登录程序
    Accessibility辅助功能--一念天堂,一念地狱
    使用FragmentTabHost+TabLayout+ViewPager实现双层嵌套Tab
    android性能优化练习:过度绘制
  • 原文地址:https://www.cnblogs.com/lzmfywz/p/3162369.html
Copyright © 2011-2022 走看看