HDU-1159: Common Subsequence（最长公共子序列）

DP

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 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 = 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

题意：求最长公共子序列长度。

分析：

• 完整代码：

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
char a[1111],b[1111];
int dp[1111][1111];//最长公共子序列长度，下标为序列ab的长度
int main()
{
    int la,lb,i,j;
    while(scanf("%s%s",a,b)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        la=strlen(a);
        lb=strlen(b);
        for(i=1;i<=la;i++)
        {
            for(j=1;j<=lb;j++)
            {
                if(a[i-1]==b[j-1])
                    dp[i][j]=dp[i-1][j-1]+1;
                else
                    dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
            }
        }
        printf("%d\n",dp[la][lb]);
    }
    return 0;
}