题意:给你n个的串,求出它们的最长公共子串,如果不存在这个子串,则输出“IDENTITY LOST”,如果存在多个最长公共子串,则输出字典序最小的那一个。
思路:枚举+KMP。与poj3080类似,详情见poj3080
#include<iostream>
#include<cstring>
using namespace std;
const int nMax = 4002;
const int mMax = 202;
char text[nMax][mMax], pat[mMax];
int n, ma, lent[nMax], lenp, next[mMax];
void get_next()
{
int j=1, k=0;
next[0] = -1;
next[1]=0;
while(j< lenp)
{
if(pat[j] == pat[k])
{
next[j+1]=k+1;
j++;
k++;
}
else if (k==0)
{
next[j+1]=0;
j++;
}
else
k=next[k];
}
}
void KMP()
{
int k, m, i, j;
get_next();
ma = 200;
for(k = 1; k < n; k ++)
{
i = 0; j = 0; m = 0;
while(i < lent[k] && j < lenp)
{
if(j == -1 || text[k][i] == pat[j])
{
i ++;
j ++;
}
else j = next[j];
if(j > m)
m = j;
}
if(m < ma) ma = m;
}
}
int main()
{
int i;
char result[mMax], tmp[mMax];
while(scanf("%d", &n) && n)
{
for(i = 0; i < n; i ++)
{
scanf("%s", text[i]);
lent[i] = strlen(text[i]);
}
int ans = 0;
for(i = 0; i < lent[0]; i ++)
{
strcpy(pat, text[0] + i);
lenp = lent[0] - i;
KMP();
if(ans < ma)
{
ans = ma;
strncpy(result, text[0] + i, ans);
result[ans] = '\0';
}
else if(ans == ma)
{
strncpy(tmp, text[0] + i, ans);
tmp[ans] = '\0';
if(strcmp(tmp, result) < 0)
strcpy(result, tmp);
}
}
if(ans == 0)
printf("IDENTITY LOST\n");
else
printf("%s\n", result);
}
return 0;
}