HDU 1159 & POJ 1458 Common Subsequence 【LCS】

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34730    Accepted Submission(s): 15856

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 abfcabprogramming contest abcd mnp

 

Sample Output

420

 

Source

Southeastern Europe 2003

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=1159

LCS(Longest  Common Subsequence)最长公共子序列模板题。

第一组测试数据的答案为：abfb.

LCS是DP的基础问题。

状态转移方程：

dp[i][j]={dp[i−1][j−1]+1,max(dp[i−1][j],dp[i][j−1]),a[i]==b[j]a[i]!=b[j]

dp[i][j]表示0到i-1跟0到j-1的最长公共子序列.

图形说明，图片来源：http://blog.csdn.net/a_eagle/article/details/7213236

AC代码1；

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn=1000+5;int dp[maxn][maxn];char a[maxn],b[maxn];int main(){    while(cin>>a>>b)    {        int len1=strlen(a);        int len2=strlen(b);        for(int i=0; i<=len1; i++)            dp[i][0]=0;        for(int i=0; i<=len2; i++)            dp[0][i]=0;        for(int i=1; i<=len1; i++)        {            for(int j=1; j<=len2; j++)            {                if(a[i-1]==b[j-1])//注意细节。                    dp[i][j]=dp[i-1][j-1]+1;                else                    dp[i][j]=max(dp[i][j-1],dp[i-1][j]);            }        }        cout<<dp[len1][len2]<<endl;    }    return 0;}

AC代码2：

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>using namespace std;const int maxn=1000+5;int dp[maxn][maxn];char a[maxn],b[maxn];int main(){    while(cin>>a+1>>b+1)    {        int len1=strlen(a+1);        int len2=strlen(b+1);        for(int i=0;i<=len1;i++)            dp[i][0]=0;        for(int i=0;i<len2;i++)            dp[0][i]=0;        for(int i=1;i<=len1;i++)        {            for(int j=1;j<=len2;j++)            {                if(a[i]==b[j])//注意细节。                    dp[i][j]=dp[i-1][j-1]+1;                else                    dp[i][j]=max(dp[i-1][j],dp[i][j-1]);            }        }        cout<<dp[len1][len2]<<endl;    }    return 0;}