The Dominator of Strings
Time Limit: 3000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 381 Accepted Submission(s): 97
Problem Description
Here you have a set of strings. A dominator is a string of the set dominating all strings else. The string S is dominated by T if S is a substring of T.
Input
The input contains several test cases and the first line provides the total number of cases.
For each test case, the first line contains an integer N indicating the size of the set.
Each of the following N lines describes a string of the set in lowercase.
The total length of strings in each case has the limit of 100000.
The limit is 30MB for the input file.
Output
For each test case, output a dominator if exist, or No if not.
Sample Input
3 10 you better worse richer poorer sickness health death faithfulness youbemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulness 5 abc cde abcde abcde bcde 3 aaaaa aaaab aaaac
【分析】：
赛时调了一下午AC自动机，一直超时。过后问ACfun群里，kmp就能过，啊啊啊，想骂人了。我一直以为kmp更超时，没敢用。
赛后用kmp的模板直接跑了一遍就过了。还是太弱了，技巧性的东西或者脑袋都不够灵活。5个小时需要灵动思路啊
【代码】：
#include <stdio.h>
#include <string.h>
char S[1200010];
char *t[1100010],*s;
int f[1202020];
void getfail(char p[],int f[]) //字符串p自我匹配
{
int len=strlen(p);
f[0]=f[1]=0;
for(int i=1;i<len;i++)
{
int j=f[i];
while(j&&p[i]!=p[j])
j=f[j];
if(p[i]==p[j])
f[i+1]=j+1;//多匹配到了一个字符
else
f[i+1]=0;//该字符配不上
}
}
int find(char* T, char*P, int*f)//p去匹配字符串T
{
int n = strlen(T), m = strlen(P);
getfail(P, f); //得出部分匹配表
int j = 0; //短串的下标
for(int i = 0; i < n; i++) //长串下标
{
while(j && P[j] != T[i])//突然失配了
{
j = f[j]; //j往回退，直到0或者上一个字符相等的位置
}
if(P[j] == T[i])
{
j++; //匹配了一个字符，j++
}
if(j == m) //短串匹配到头了
{
return 1;//i - m + 1;//返回成功匹配的起点字符位置
}
}
return -1;
}
int main()
{
int T,n;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
int maxlen=0;
int p=1;//记录最长串
s=S;
for(int i=1;i<=n;i++)
{
scanf("%s",s);
t[i]=s;
if(strlen(s)>maxlen){
maxlen=strlen(s);
p=i;
}
s+=strlen(s)+2;
}
int ans=0;
for(int i=1;i<=n;i++)
{
if(find(t[p],t[i],f)==1)
ans++;
else break;
}
if(ans==n)
{
printf("%s\n",t[p]);
}
else puts("No");
}
return 0;
}
