最长公共子序列

时间：2017-07-25 11:14:40 阅读：226 评论：0 收藏：0 [点我收藏+]

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9595 Accepted Submission(s): 3923

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

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

一个简单的动态规划的应用

直接给代码

 1 #include <iostream>
 2 //#include <cstdio>
 3 #include <cstring>
 4 #define N 1005
 5 using namespace std;
 6 int dp[N][N];
 7 string s,ss;
 8 int main(){
 9     while(cin>>s>>ss){
10         int n=s.length();
11         int m=ss.length();
12         memset(dp,0, sizeof(dp));
13         for(int i=1;i<=n;i++){
14             for(int j=1;j<=m;j++){
15                 if(s[i-1]==ss[j-1])
16                     dp[i][j]=dp[i-1][j-1]+1;
17                 else{
18                     dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
19                 }
20             }
21         }
22         cout<<dp[n][m]<<endl;
23     }
24     return 0;
25 }

最长公共子序列

原文：http://www.cnblogs.com/zllwxm123/p/7232930.html

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