标签:out closed define 题意 sign pre cin std ble
题目链接: http://acm.split.hdu.edu.cn/showproblem.php?pid=6208
题意:给你n个串,能不能在这n个串中找到一个串,使得其它所有的串都是这个串的子串。
题解:找到最长的串作为主串,将主串hash一下,然后去匹配其它串即可
代码:
1 #pragma GCC diagnostic error "-std=c++11" 2 #include <bits/stdc++.h> 3 using namespace std; 4 #define mem(a,b) memset(a,b,sizeof(a)) 5 #define FIN freopen("in.txt","r",stdin) 6 #define IO ios_base::sync_with_stdio(0),cin.tie(0) 7 #define pb push_back 8 typedef long long LL; 9 typedef unsigned long long ULL; 10 typedef pair<int, int> PIR; 11 12 const int MAXN = 1e5+5; 13 14 int T, n; 15 ULL H[MAXN], qpow[MAXN]; 16 string S, s[MAXN]; 17 map <ULL, bool> vis; 18 19 void pre () { 20 qpow[0] = 1; 21 for (int i = 1; i <= 1e5; ++i) qpow[i] = qpow[i-1]*27; 22 } 23 ULL getHash (int l, int r) { 24 if (l == 0) return H[r]; 25 else return (H[r]-H[l-1]*qpow[r-l+1]); 26 } 27 void solve () { 28 H[0] = S[0]-‘a‘+1; 29 for (int i = 1; i < S.size(); ++i) H[i] = H[i-1]*27+S[i]-‘a‘+1; 30 31 for (int i = 1; i <= n; ++i) { 32 if (s[i] == S) continue; 33 ULL e = s[i][0]-‘a‘+1, e1 = S[0]-‘a‘+1; 34 for (int j = 1; j < s[i].size(); ++j) e = e*27+s[i][j]-‘a‘+1; 35 int ok = 0; 36 for (int j = 0; j < S.size(); ++j) { 37 if (j+s[i].size()-1 >= S.size()) break; 38 ULL x = getHash(j, j+s[i].size()-1); 39 if (x == e) { 40 ok = 1; 41 break; 42 } 43 } 44 if (!ok) { 45 cout << "No" << endl; 46 return ; 47 } 48 } 49 cout << S << endl; 50 } 51 int main () 52 {IO; 53 //FIN; 54 pre(); 55 cin >> T; 56 while (T--) { 57 vis.clear(); 58 cin >> n; 59 int maxn = 0, id = -1; 60 for (int i = 1; i <= n; ++i) { 61 cin >> s[i]; 62 if (s[i].size() > maxn) { 63 maxn = s[i].size(); 64 id = i; 65 } 66 } 67 S = s[id]; 68 solve(); 69 } 70 return 0; 71 }
标签:out closed define 题意 sign pre cin std ble
原文地址:http://www.cnblogs.com/Jstyle-continue/p/7647368.html