hdoj 1159 Common Subsequence【LCS】【DP】

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28494    Accepted Submission(s): 12735

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

 

非常基础的一道LCS。以下给出第一种情况的DP路线，例如以下图。

希望能够帮助大家。

AC代码：

#include<stdio.h>
#include<string.h>
#define max(a,b) (a>b?
a:b)
char a[1000],s[1000];
int dp[1000][1000];
int main()
{
    int i,j,k;
    while(scanf("%s%s",a,s)!=EOF)
    {
    	memset(dp,0,sizeof(dp));
        int l=strlen(a);
        int le=strlen(s);
        for(i=1;i<=l;i++)
        {
            for(j=1;j<=le;j++)
            if(a[i-1]==s[j-1])//推断左側和上側字符是否相等 
                dp[i][j]=dp[i-1][j-1]+1;//把左上側的dp值+1 
            else
                dp[i][j]=max(dp[i-1][j],dp[i][j-1]);//取左側或上側的最大dp值    
        }
        printf("%d
",dp[l][le]);
    }
    return 0;
}