HDU 6208 The Dominator of Strings 后缀自动机

The Dominator of Strings

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

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：

　　给你n个串，问你是否村你在一个串，包含了其他所有串

题解：

　　必然是最长串，

　　将 其建立后缀自动机， 并将剩余的串与其匹配，也就是最长公共字串 长度　　

　　ios::sync_with_stdio(false); cin.tie(0); 这句话 可以加快C++输入输出，不加就TLE

#include <bits/stdc++.h>
inline long long read(){long long x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;}
using namespace std;

const int N = 2e5+7;
const long long mod = 1000000007;

int isPlus[N * 2],endpos[N * 2];int d[N * 2];
int tot,slink[2*N],trans[2*N][28],minlen[2*N],maxlen[2*N],pre;
int newstate(int _maxlen,int _minlen,int* _trans,int _slink){
    maxlen[++tot]=_maxlen;minlen[tot]=_minlen;
    slink[tot]=_slink;
    if(_trans)for(int i=0;i<26;i++)trans[tot][i]=_trans[i];
    return tot;
}
int add_char(char ch,int u){
    int c=ch-'a',v=u;
    int z=newstate(maxlen[u]+1,-1,NULL,0);
    isPlus[z] = 1;
    while(v&&!trans[v][c]){trans[v][c]=z;d[z]+=1;v=slink[v];}
    if(!v){ minlen[z]=1;slink[z]=1;return z;}
    int x=trans[v][c];
    if(maxlen[v]+1==maxlen[x]){slink[z]=x;minlen[z]=maxlen[x]+1;return z;}
    int y=newstate(maxlen[v]+1,-1,trans[x],slink[x]);
    slink[z]=slink[x]=y;minlen[x]=minlen[z]=maxlen[y]+1;
    while(v&&trans[v][c]==x){trans[v][c]=y;d[x]--,d[y]++;v=slink[v];}
    minlen[y]=maxlen[slink[y]]+1;
    return z;
}
void init_sam() {
    for(int i = 1; i <= tot; ++i)
        for(int j = 0; j < 26; ++j) trans[i][j] = 0;
    pre = tot = 1;
}
int T,n;
string a[N];
int main() {
    ios::sync_with_stdio(false); cin.tie(0);
    cin>>T;
    while(T--) {
        cin>>n;
        int mx = 0,flag = 0;int ok = 0;
        for(int i = 1; i <= n; ++i) {
            cin >> a[i];
            if(a[i].length() > mx) {
                mx = a[i].length();
                flag = i;
            }
        }
        init_sam();
        for(int i = 0; i < mx; ++i)
            pre = add_char(a[flag][i],pre);
        for(int i = 1; i <= n; ++i) {
            if(i == flag) continue;
            int now = 0;
            int p = 1;
            int ans = 0;
            for(int j = 0; j < a[i].length(); ++j) {
                if(trans[p][a[i][j]-'a']) {
                    now += 1;
                    p = trans[p][a[i][j]-'a'];
                }else {
                    while(p) {
                        p = slink[p];
                        if(trans[p][a[i][j]-'a']) break;
                    }
                    if(!p) p = 1,now = 0;
                    else {
                        now = maxlen[p] + 1;
                        p = trans[p][a[i][j]-'a'];
                    }
                }
                ans = max(ans,now);
            }
            if(ans != a[i].length()) {
                cout<<"No"<<endl;
                ok = 1;
                break;
            }
        }
        if(!ok) cout<<a[flag]<<endl;
    }
    return 0;
}