HDU–6208 The Dominator of Strings[KMP]

The Dominator of Strings

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

题意：所给的所有字符串中是否含有一个主串包含其他的字符串。

思路：直接找到最长的，检查。

#include "bits/stdc++.h"
using namespace std;
const int maxn = 2000000 + 10;
int f[maxn];
int T,N;
char s[maxn];
char *t[maxn];
void getfail(char p[]) {
    int len = strlen(p);  
    f[0] = f[1] = 0;  
    for(int i = 1; i < len; i++) {
        int j = f[i];  
        while(j and p[i] != p[j]) j = f[j];  
        if(p[i] == p[j]) f[i + 1] = j + 1; 
        else f[i + 1] = 0;   
    }   
}  
int find(char T[], char P[]) {
    int n = strlen(T), m = strlen(P);    
    getfail(P); int j = 0;  
    for(int i = 0; i < n; i++) {
        while(j && P[j] != T[i]) j = f[j];    
        if(P[j] == T[i]) j++; 
        if(j == m) return 1;
    }    
    return -1;    
} 
int main(int argc, char const *argv[])
{
    scanf("%d",&T);
    while(T--){
        scanf("%d",&N);
        int maxx = 0;
        int tlen = 0;
        char *ms, *ts=s;
        for(int i=1 ;i <= N;i++){
            scanf("%s",ts);
            tlen = strlen(ts);
            if(tlen > maxx){
                maxx = tlen; ms = ts;
            }
            t[i] = ts; ts += strlen(ts) + 2;
        }
        bool flag = true;
        for(int j = 1;j <= N; j++){
            if( find(ms,t[j]) != 1 ){
                flag = false; break;
            }
        }
        if(flag) printf("%s
",ms);
        else printf("No
");
    }
    return 0;
}