HDUOJ--1159Common Subsequence
Common Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 19595 Accepted Submission(s): 8326
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
Source
简单的lcs
lcs 模板：
int lcs(const char *a,const char *b)
{
int i,j;
int m=strlen(a),n=strlen(b);
mar[0][0]=0;
for(i=1;i<=m;i++)
mar[i][0]=0;
for(i=1;i<=n;i++)
mar[0][i]=0;
for(i=1;i<=m;i++)
{
for(j=1;j<=n;j++)
{
if(a[i-1]==b[j-1])
mar[i][j]=mar[i-1][j-1]+1;
else
mar[i][j]=mar[i-1][j]>mar[i][j-1]?mar[i-1][j]:mar[i][j-1];
}
}
return mar[m][n];
}此题的代码：
1 #include<stdio.h>
2 #include<string.h>
3 #define maxn 1000
4 int mar[maxn][maxn];
5 char x[maxn],y[maxn];
6
7 int lcs(const char *a,const char *b)
8 {
9 int i,j;
10 int m=strlen(a),n=strlen(b);
11 mar[0][0]=0;
12 for(i=1;i<=m;i++)
13 mar[i][0]=0;
14 for(i=1;i<=n;i++)
15 mar[0][i]=0;
16 for(i=1;i<=m;i++)
17 {
18 for(j=1;j<=n;j++)
19 {
20 if(a[i-1]==b[j-1])
21 mar[i][j]=mar[i-1][j-1]+1;
22 else
23 mar[i][j]=mar[i-1][j]>mar[i][j-1]?mar[i-1][j]:mar[i][j-1];
24 }
25 }
26 return mar[m][n];
27 }
28 int main()
29 {
30 while(scanf("%s%s",&x,&y)!=EOF)
31 printf("%d\n",lcs(x,y));
32 return 0;
33 }