HDU 6208 The Dominator of Strings 后缀自动机

最新推荐文章于 2025-08-08 22:01:10 发布

weixin_30699463

最新推荐文章于 2025-08-08 22:01:10 发布

阅读量61

点赞数

CC 4.0 BY-SA版权

文章标签： java c/c++ 数据结构与算法

原文链接：http://www.cnblogs.com/zxhl/p/7617784.html

本文介绍了一种寻找能支配所有其他字符串的主导字符串的方法。通过构建后缀自动机并进行匹配，快速判断是否存在一个字符串包含集合中所有其他字符串。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

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

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;
}