POJ 1226 Substrings(最长公共连续串的变形,可以倒转+KMP)
Substrings
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 10451 | Accepted: 3598 |
Description
You are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings.
Input
The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains a single integer n (1 <= n <= 100), the number of given strings, followed by n lines, each representing one string of minimum length 1 and maximum length 100. There is no extra white space before and after a string.
Output
There should be one line per test case containing the length of the largest string found.
Sample Input
2 3 ABCD BCDFF BRCD 2 rose orchid
Sample Output
2 2
题目大意：给你n个串，求最小公共连续子串，不过这个可以倒转，比如ro和or可以匹配
解题思路：思路很清晰，把第一个串当模式串，分解求next匹配然后逆转求next匹配。自己的思路是先顺序找到子串，然后再找它的逆转串的相对位置。例如模式串abcdefg,我先逆转模式串放在另一个数组里面。子串可以是bcde（sta=1,len=1+4）逆转就变成了edcb(sta=7-1-4,len=7-1).这样把规律看出来就可以了。果断一A!
题目地址：Substrings
AC代码：
#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
using namespace std;
int nextp[105],nexts[105],n;
char a[105][105];
char b[105]; //将模式串翻转
int lena0; //模式串的长度
//我的思路是用第一个字符串当模式串，然后再枚举长度由大到小求next匹配
void getnextp(int sta,int len) //模式串
{
int i,j;
len=sta+len; //相对位置变成绝对位置
nextp[sta]=sta,nextp[sta+1]=sta;
for(i=sta+1;i<len;i++)
{
j=nextp[i];
while(j!=sta&&a[0][i]!=a[0][j])
j=nextp[j];
if(a[0][i]==a[0][j])
nextp[i+1]=j+1;
else
nextp[i+1]=sta;
}
}
void getnexts(int sta,int len) //翻转后的模式串
{
int i,j;
sta=lena0-sta-len; //要把子串的位置转换成b的位置
len=sta+len;
nexts[sta]=sta,nexts[sta+1]=sta;
for(i=sta+1;i<len;i++)
{
j=nexts[i];
while(j!=sta&&b[i]!=b[j])
j=nexts[j];
if(b[i]==b[j])
nexts[i+1]=j+1;
else
nexts[i+1]=sta;
}
}
int KMP(int sta,int len)
{
int i,j,k;
int sta1=sta;
int sta2=lena0-sta-len;
int len1=sta+len;
int len2=sta2+len;
for(k=1;k<n;k++) //后面的串都和第一个串的子串匹配
{
int flag=0;
j=sta1;
for(i=0;i<strlen(a[k]);i++)
{
while(j!=sta1&&a[k][i]!=a[0][j])
j=nextp[j];
if(a[k][i]==a[0][j])
j++;
if(j==len1)
{
flag=1;
break;
}
}
j=sta2;
for(i=0;i<strlen(a[k]);i++)
{
while(j!=sta2&&a[k][i]!=b[j])
j=nexts[j];
if(a[k][i]==b[j])
j++;
if(j==len2)
{
flag=1;
break;
}
}
if(flag==0) //有一个没有匹配，便返回0
return 0;
}
return 1;
}
int main()
{
int i,j,T,l;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
if(n==0)
break;
for(i=0;i<n;i++)
scanf("%s",a[i]);
lena0=strlen(a[0]);
for(i=0;i<lena0;i++)
b[i]=a[0][lena0-i-1];
b[lena0]='\0';
int flag=0;
for(i=lena0;i>=1;i--) //从长度lena0开始往下枚举
{
int flag1=0; //长度为i的是否存在
for(j=0;j<=lena0-i;j++) //模式串分解的起始位置
{
getnextp(j,i);
getnexts(j,i);
if(KMP(j,i))
{
flag1=1;
printf("%d\n",i);
break;
}
}
if(flag1)
{
flag=1;
break;
}
}
if(flag==0)
puts("0");
}
return 0;
}