标签：des style blog http color io os ar java
Common Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23973 Accepted Submission(s): 10592
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
没什么说的、上代码
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define max(a,b) a>b?a:b
#define INF 0x7ffffff
#define N 1010
int dp[N][N]; //dp[i][j]表示a中前i个字符和b中前j个字符的最长公共子序列
char s1[N],s2[N];
int main()
{
int i,j;
while(scanf("%s%s",s1+1,s2+1)!=EOF)
{
int l1=strlen(s1+1);
int l2=strlen(s2+1);
memset(dp,0,sizeof(dp));
for(i=1;i<=l1;i++)
{
for(j=1;j<=l2;j++)
{
if(s1[i]==s2[j])
{
dp[i][j]=dp[i-1][j-1]+1;
}
else
{
dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
}
}
}
cout<<dp[l1][l2]<<endl;
}
return 0;
}
最长公共子序列 [HDU 1159] Common Subsequence
标签：des style blog http color io os ar java
原文地址：http://www.cnblogs.com/hate13/p/4050418.html
