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 題意：給定一系列的字串，讓找出一個字串，可以包含其他所有的字串。

思路：找出最長的那個字串，對它進行hash，它一定是可能滿足條件的，如果對於最長的長度有多個字串，如果存在滿足條件的字串，那麼這幾個等長的字串一定相同。對於其他的字串，判斷最長的字串中是否含有其他字串就行了。（用最原始的暴力就能過， 可能是因為沒有大的資料）。

程式碼：

#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[N], p[N];
char str[N];
char temp[N];
ull value[N];
int length[N];
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;
}