杭电1159-Common Subsequence

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17428    Accepted Submission(s): 7316

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
一道求最长公共子序列的题目，刚开始不知道是用DP求后来看了别人的代码才知道这是用一道简单的DP题目，我当时顿时无语了
AC代码：
#include<iostream>//最长公共子序列
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
const int MAX=1001;
char x[MAX];
char y[MAX];
int DP[MAX][MAX];
using namespace std;
int Max(int x,int y)
{
    return x>y?x:y;
}
int main()
{
    int n,m,i,j;
    while(cin>>x>>y)
    {
        n=strlen(x);
        m=strlen(y);
        memset(DP,0,sizeof(DP));
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            {
                if(x[i-1]==y[j-1])
                DP[i][j]=DP[i-1][j-1]+1;//如果找到了就在前一步加+1
                else
                DP[i][j]=Max(DP[i-1][j],DP[i][j-1]);//没有的就看哪个序列最长
            }
        }
        cout<<DP[n][m]<<endl;
    }
    return 0;
}