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
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.
int lcs(const char *a,const char *b)
{
int i,j;
int m=strlen(a),n=strlen(b);
mar[][]=;
for(i=;i<=m;i++)
mar[i][]=;
for(i=;i<=n;i++)
mar[][i]=;
for(i=;i<=m;i++)
{
for(j=;j<=n;j++)
{
if(a[i-]==b[j-])
mar[i][j]=mar[i-][j-]+;
else
mar[i][j]=mar[i-][j]>mar[i][j-]?mar[i-][j]:mar[i][j-];
}
}
return mar[m][n];
}
此题的代码:
#include<stdio.h>
#include<string.h>
#define maxn 1000
int mar[maxn][maxn];
char x[maxn],y[maxn]; int lcs(const char *a,const char *b)
{
int i,j;
int m=strlen(a),n=strlen(b);
mar[][]=;
for(i=;i<=m;i++)
mar[i][]=;
for(i=;i<=n;i++)
mar[][i]=;
for(i=;i<=m;i++)
{
for(j=;j<=n;j++)
{
if(a[i-]==b[j-])
mar[i][j]=mar[i-][j-]+;
else
mar[i][j]=mar[i-][j]>mar[i][j-]?mar[i-][j]:mar[i][j-];
}
}
return mar[m][n];
}
int main()
{
while(scanf("%s%s",&x,&y)!=EOF)
printf("%d\n",lcs(x,y));
return ;
}
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