hdu 6208 hash字符串水题

The Dominator of Strings

Time Limit: 3000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 2694    Accepted Submission(s): 975


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

 

Sample Output

youbemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulness
abcde
No

 

Source

2017 ACM/ICPC Asia Regional Qingdao Online
题意：给定一系列的字符串，让找出一个字符串，可以包含其他所有的字符串。

思路：找出最长的那个字符串，对它进行hash，它一定是可能满足条件的，如果对于最长的长度有多个字符串，如果存在满足条件的字符串，那么这几个等长的字符串一定相同。对于其他的字符串，判断最长的字符串中是否含有其他字符串就行了。（用最原始的暴力就能过， 可能是因为没有大的数据）。

代码：

#include <stdio.h>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include<queue>
#include<iostream>
#include<list>
using namespace std;
typedef unsigned long long ull;
const int N = 1e5+5;
const int X = 163;
ull hhash
, p
;
char str
;
char temp
;
ull value
;
int length
;
int m;
void inIt() {
p[0] = 1;
for(int i = 1; i <= N - 5; i++)
p[i] = p[i-1] * X;
}

ull get(int l, int r, ull g[]) {
return g[r] - g[l-1]*p[r-l+1];
}

int main() {
inIt();
int t, index, Max, flag, len;
ull sum;
scanf("%d", &t);
while(t--) {
scanf("%d", &m);
Max = 0;
for(int i = 1; i <= m; i++) {
scanf("%s", str);
len = strlen(str);
if(len > Max) {
Max = len;
strcpy(temp, str);
}
sum = 0;
for(int j = 1; j <= len; j++)
sum = sum * X + (str[j-1] - 'a');
value[i] = sum;
length[i] = len;
}
hhash[0] = 0;
for(int i = 1; i <= Max; i++) {
hhash[i] = hhash[i-1]*X + (temp[i-1] - 'a');
}

for(int i = 1; i <= m; i++) {
flag = 0;
len = length[i];
for(int k = 1; k <= Max-len+1; k++) {
if(value[i] == get(k, k+len-1, hhash)) {
flag = 1;
break;
}
}
if(!flag) {
break;
}
}
if(flag) {
printf("%s\n", temp);
}
else
printf("No\n");
}
return 0;
}