EOJ382 Match Maker
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submits: 50 | Accepted: 6 |
Description
In Computer Science, pattern matching is the act of checking if a certain sequence conforms (matches) a given pattern. Patterns are usually specified using a language based on regular expression. In this problem, we'll use a simple regular expression to express patterns on sequences of decimal digits. A pattern is a sequence of one or more decimal digits `0' ...`9', asterisks `*', and hash signs `#'. A `*' denotes a sequence of an even number of digits, whereas a `#' denotes a sequence of an odd number of digits. For example, the pattern ``129" only matches the sequence 129. The pattern ``1*3" matches all sequences beginning with 1, ending with 3, and having an even number of decimal digits between the first and last digits. As another example, the pattern ``#55" matches the sequences 155, 12355, 1234555, but none of the sequences 55, 1255, 123455. Your task is to write a program to find if a given sequence matches a given pattern.
Input
Your program will be tested on one or more data sets. Each data set contains a single pattern and one or more sequences to match. The first line of each data set specifies the pattern, and the remaining lines specify the sequences to match against that pattern. The end of a data set (except the last) is identified by the word ``END" (without the double quotes.) The end of the last data set is identified by the word ``QUIT". All lines are 100,000 characters long or shorter.
Output
k.s. result
Where k is the test case number (starting at one,) and s is the sequence number (starting at one within each test case,) and result is either the word ``match" if the given string matches the pattern, or the word ``not" if it doesn't.
Where k is the test case number (starting at one,) and s is the sequence number (starting at one within each test case,) and result is either the word ``match" if the given string matches the pattern, or the word ``not" if it doesn't.
Sample Input
129
1299
129
1129
END
1*3
123
1223
END
#55
155
12355
55
1255
QUIT
1299
129
1129
END
1*3
123
1223
END
#55
155
12355
55
1255
QUIT
Sample Output
1.1. not
1.2. match
1.3. not
2.1. not
2.2. match
3.1. match
3.2. match
3.3. not
3.4. not
1.2. match
1.3. not
2.1. not
2.2. match
3.1. match
3.2. match
3.3. not
3.4. not
*************************************************
题目大意：字符串匹配，一个含有数字、'#'、'*'的模式串去匹配一个纯数字的字符串，一个'#'可以匹配奇数个数字，一个'*'可以匹配偶数个数字，如果可以匹配就输出match否则就输出not。
解题思路：根据'#'、'*'把模式串分成几个子模式串进行KMP。
以下是我的代码，又长又搓，哎。。
#include <stdio.h>
#include <string.h>
#define N 100005
char a[N],b[N],c[N];
int ed,lf[N],p[N],l1,l2,s[N],e[N],top;
int solve(int s,int e,int kf)
{
int j=-1;
for(int i=ed+1;i<l2;i++)
{
while(1)
{
if(a[j+1+s]==b[i])
{
j++;
if(j+s+1==e)
{
if(((i-ed-e+s)&1)==kf)
{
ed=i;
return 1;
}
else
{
if(j==0)break;
j--;
j=p[j+s];
continue;
}
}
break;
}
if(j==-1)break;
j=p[j+s];
}
}
return 0;
}
void run(void)
{
int t1=0,t2=0;
while(1)
{
if(!t2)
{
t1++;t2=1;
scanf("%s",a+1);
if(a[1]=='Q')return ;
if(a[1]=='E')
{
t2=0;
continue;
}
a[0]='0';
l1=strlen(a);
a[l1++]='0';
memset(lf,0,sizeof(lf));
memset(p,-1,sizeof(p));
top=0;
int i=0;
for(i=0;i<l1&&a[i]>='0'&&a[i]<='9';i++);
int flag=0;
for(i;i<l1;i++)
{
if(a[i]>='0'&&a[i]<='9')
{
if(!flag)
s[++top]=i;
flag=1;
}
else
{
if(flag)
{
e[top]=i;
for(int i=s[top]+1;i<e[top];i++)
{
int k=p[i-1];
while(1)
{
if(a[k+1+s[top]]==a[i])
{
p[i]=k+1;
break;
}
if(k==-1)break;
k=p[k+s[top]];
}
}
}
flag=0;
if(a[i]=='#')
lf[top+1]^=1;
}
}
if(top)e[top]=l1;
continue;
}
else
{
t2++;
scanf("%s",b+1);
if(b[1]=='Q')return ;
if(b[1]=='E')
{
t2=0;
continue;
}
b[0]='0';
l2=strlen(b);
b[l2++]='0';
}
int i,flag=0;
for(i=0;i<l1&&a[i]>='0'&&a[i]<='9'&&!flag;i++)
if(a[i]!=b[i])
flag=1;
if(flag||(i==l1&&i<l2))
{
printf("%d.%d. not\n",t1,t2-1);
continue;
}
if(i==l1&&i==l2)
{
printf("%d.%d. match\n",t1,t2-1);
continue;
}
ed=i-1;
flag=0;
int mark=0;
for(int i=1;i<top&&!mark;i++)
if(solve(s[i],e[i],lf[i])==0)
mark=1;
if(mark)
{
printf("%d.%d. not\n",t1,t2-1);
continue;
}
int k=l2-1;
for(int j=l1-1;j>=0&&a[j]>='0'&&a[j]<='9';j--)
if(a[j]!=b[k--])
mark=1;
if(mark||k<ed||((k-ed)&1)!=lf[top])
{
printf("%d.%d. not\n",t1,t2-1);
continue;
}
printf("%d.%d. match\n",t1,t2-1);
}
}
int main()
{
run();
return 0;
}