HDU 1159 Common Subsequence （dp）

最新推荐文章于 2021-07-13 14:09:30 发布

原创最新推荐文章于 2021-07-13 14:09:30 发布 · 283 阅读

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本文探讨了最长公共子序列问题的解决方法，通过对比两种不同的动态规划算法实现方式，展示了如何寻找两个字符串间的最长公共子序列长度。文章提供了具体的代码示例及样例输入输出。

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Common Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42749 Accepted Submission(s): 19708

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

用线性dp发现会出现错误 aa a两个数列答案是 2

错误代码:

#include <stdio.h>
#include<algorithm>
#include<string.h> 
using namespace std;
int main(int argc, char *argv[])
{
    char a[1001],b[1001];
    while(scanf("%s %s",a,b)!=EOF)
    {
        int dp[1001]={0};
        int i,j,k;
        int len1=strlen(a);
        int len2=strlen(b);
        for(i=0;i<len2;i++)
        {
            for(j=0;j<len1;j++)
            {
                if(b[i]==a[j])
                {
                    dp[j]=1;
                    for(k=j-1;k>=0;k--)
                    {
                        dp[j]=max(dp[j],dp[k]+1);
                    }
                }
            } 
        }
        int max1=0;
        for(i=0;i<len1;i++)
        {
            max1=max(dp[i],max1);
        }
        printf("%d\n",max1);
    }
    return 0;
}

所以用矩阵类dp

#include <stdio.h>
#include<string.h>
int dp[1010][1010];
int main(int argc, char *argv[])
{
int i,j;
char a[1000],b[1000];
while(scanf("%s %s",a,b)!=EOF)
    {
    int len1,len2;
    len1=strlen(a);
    len2=strlen(b);
    for(i=0;i<=len1;i++)
    {
    dp[0][i]=0;    
    }
    for(i=0;i<=len2;i++)
    {
        dp[i][0]=0;
    }    
    for(i=1;i<=len1;i++)
    {
        for(j=1;j<=len2;j++)
        {
        if(a[i-1]==b[j-1])
        {
        dp[i][j]=dp[i-1][j-1]+1;    
        }    
        else
        {
            if(dp[i][j-1]>dp[i-1][j])
            {
                dp[i][j]=dp[i][j-1];
            }
            else
            {
                dp[i][j]=dp[i-1][j];
            }
        }
        }
    }
    printf("%d\n",dp[len1][len2]);
    }
    return 0;
}