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)
Total Submission(s): 1369    Accepted Submission(s): 475

Problem Description

Here you have a set of strings. A dominator is a string of the set dominating all strings else. The stringS 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

310youbetterworsericherpoorersicknesshealthdeathfaithfulnessyoubemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulness5abccdeabcdeabcdebcde3aaaaaaaaabaaaac

 

Sample Output

youbemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulnessabcdeNo

 

Source

2017 ACM/ICPC Asia Regional Qingdao Online 

HDU 6208 The Dominator of Strings

My Solution

题意：每组数据给出n个字符串，每组总共最多1e5个字符，然后要求判断，其中是否存在一个字符串，而其它字符串是这个字符串的子串。总共有30MB的输入。

贪心+后缀自动机

如果用AC自动机来做很可能会TLE，但后缀自动机在这里相对不容易被卡掉，首先我们多出了一个贪心优化，那就是那个可能存在的母串必定是长度最长的字符串，如果长度最长的字符串不止一个，则它们必定是完全相同的，所以后缀自动机只会用最长的那个字符串来建立自动机，而如果是AC自动机则必须使用全部的字符串来建立自动机，加上乘个memset[26]很容易被卡掉，如果非要卡后缀自动机可以把这30MB的输入数据都处理为每组数据2个字符串，一个长度为1e5-1一个长度为1，这样每次都要用最长串来建立自动机，所以复杂度为O(26*30M) == 7.8e8，在加上sam来做常数很小，所以这里要卡后缀自动机相对麻烦写，然后写了一发后缀自动机就过了。

首先找出最长的一个字符串建立后缀自动机，然后由于sam存放的是所以的后缀，然后后缀的前缀就是子串，所以只要对剩下的n-1个字符串分别从sam的0节点开始向下跑即可，如果能跑完就说是母串的某个后缀的前缀。

时间复杂度 O(n)

空间复杂度 O(2*n*26)

#include <iostream>#include <cstdio>#include <string>#include <cstring>using namespace std;typedef long long LL;const int MAXN = 2*1e5 + 8;string s;struct SAM{    int ch[MAXN][26], pre[MAXN], val[MAXN];    int last, tot, minl[MAXN], maxl[MAXN];    void init(){        last = tot = 0;        memset(ch[0], -1, sizeof ch[0]);        pre[0] = -1; val[0] = 0;        minl[0] = maxl[0] = 0;    }    void extend(int c){        int p = last, np = ++tot;        val[np] = val[p] + 1;        memset(ch[np], -1, sizeof ch[np]);        minl[np] = maxl[np] = val[np];        while(~p && ch[p][c] == -1) ch[p][c] = np, p = pre[p];        if(p == -1) pre[np] = 0;        else{            int q = ch[p][c];            if(val[q] != val[p] + 1){                int nq = ++tot;                memcpy(ch[nq], ch[q], sizeof ch[q]);                val[nq] = val[p] + 1;                minl[nq] = maxl[nq] = val[nq];                pre[nq] = pre[q];                pre[q] = pre[np] = nq;                while(~p && ch[p][c] == q) ch[p][c] = nq, p = pre[p];            }            else pre[np] = q;        }        last = np;    }    bool _find(const string &s){        int sz = s.size(), u = 0, c, i;        for(i = 0; i < sz; i++){            c = s[i] - 'a';            if(~ch[u][c]) u = ch[u][c];            else{                return false;            }        }        return true;    }} sam;string str[MAXN];int main(){    #ifdef LOCAL    freopen("a.txt", "r", stdin);    //freopen("a.out", "w", stdout);    #endif // LOCAL    ios::sync_with_stdio(false); cin.tie(0);    int T, n, sz, i, len, ind;    bool ans;    cin >> T;    while(T--){        cin >> n;        len = 0;        for(i = 0; i < n; i++){            cin >> str[i];            if(len < str[i].size()){                len = str[i].size();                ind = i;            }        }        s.clear();        s += str[ind];        sz = s.size();        sam.init();        for(i = 0; i < sz; i++) sam.extend(s[i] - 'a');        ans = true;        for(i = 0; i < n; i++){            if(i == ind) continue;            if(!sam._find(str[i])){                ans = false;                break;            }        }        if(ans) cout << str[ind] << "\n";        else cout << "No\n";    }    return 0;}

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