acm 3 1002 子序列

1.1002

2.                                                                                                                                    Problem B 

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 166   Accepted Submission(s) : 71

Problem Description
A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = &lt;x1, x2, ..., xm&gt; another sequence Z = &lt;z1, z2, ..., zk&gt; is a subsequence of X if there exists a strictly increasing sequence &lt;i1, i2, ..., ik&gt; of indices of X such that for all j = 1,2,...,k, xij = zj. For example, Z = &lt;a, b, f, c&gt; is a subsequence of X = &lt;a, b, c, f, b, c&gt; with index sequence &lt;1, 2, 4, 6&gt;. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y. <br>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. <br>
 
Sample Input
abcfbc abfcab
programming contest 
abcd mnp
 
Sample Output
4
2
0

3.典型的动态规划问题

4.求最长公共子序列

5.

#include <stdio.h>
#include<iostream>
#include<cstdio>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[1005][1005];
int main()
{
    int n,i,j;
    string s1,s2;
    char arr[1005];


    while(~scanf("%s",arr))
    {
        s1=arr;
        scanf("%s",arr);
        s2=arr;
        memset(dp,0,sizeof(dp));


        for(i=0;i<s1.length();i++)
        {
            for(j=0;j<s2.length();j++)
            {
                if(s1[i]==s2[j])
                {
                    dp[i+1][j+1]=dp[i][j]+1;
                }
                else
                {
                    dp[i+1][j+1]=max(dp[i][j+1],dp[i+1][j]);
                }
            }
        }
        cout<<dp[s1.length()][s2.length()]<<endl;


    }
}