您的位置:首页 > 产品设计 > UI/UE

poj 1458 Common Subsequence

2015-08-13 20:11 417 查看
[align=center]Common Subsequence[/align]

Time Limit: 1000MSMemory Limit: 10000K
Total Submissions: 43303Accepted: 17580
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.

Input
The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input
data are correct.

Output
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

Source
Southeastern Europe 2003

题意:

输入两个字符串,让你求最长公共子序列(这个子序列可以不连续),所以用LCS算法很容易就能解决。虽然讲的时候有点麻烦,但是代码非常简单,看代码吧!

代码:

#include <stdio.h>
#include <string.h>
#define max(a,b) (a>b?a:b)
int n;
char a[1005],b[1005];
int p[1005][1005];
int main()
{
int len1,len2;
while(scanf("%s%s",a,b)!=EOF)
{
len1=strlen(a);
len2=strlen(b);
for(int i=1;i<=len1;i++)
for(int j=1;j<=len2;j++)
{
if(a[i-1]==b[j-1])
{
p[i][j]=p[i-1][j-1]+1;
}
else
{
p[i][j]=max(p[i-1][j],p[i][j-1]);
}
}
printf("%d\n",p[len1][len2]);
}
return 0;
}


思路:

搜索
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: