方法:AC自动机
题意比较明显,用ac 自动机。陷阱是可能会出现重复的string。
code:
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <iostream> 5 #include <string> 6 #include <vector> 7 #include <stack> 8 #include <bitset> 9 #include <cstdlib> 10 #include <cmath> 11 #include <set> 12 #include <list> 13 #include <deque> 14 #include <map> 15 #include <queue> 16 #include <fstream> 17 #include <cassert> 18 #include <unordered_map> 19 #include <unordered_set> 20 #include <cmath> 21 #include <sstream> 22 #include <time.h> 23 #include <complex> 24 #include <iomanip> 25 #define Max(a,b) ((a)>(b)?(a):(b)) 26 #define Min(a,b) ((a)<(b)?(a):(b)) 27 #define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a)) 28 #define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a)) 29 #define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a)) 30 #define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a)) 31 #define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a)) 32 #define FOREACH(a,b) for (auto &(a) : (b)) 33 #define rep(i,n) FOR(i,0,n) 34 #define repn(i,n) FORN(i,1,n) 35 #define drep(i,n) DFOR(i,n-1,0) 36 #define drepn(i,n) DFOR(i,n,1) 37 #define MAX(a,b) a = Max(a,b) 38 #define MIN(a,b) a = Min(a,b) 39 #define SQR(x) ((LL)(x) * (x)) 40 #define Reset(a,b) memset(a,b,sizeof(a)) 41 #define fi first 42 #define se second 43 #define mp make_pair 44 #define pb push_back 45 #define all(v) v.begin(),v.end() 46 #define ALLA(arr,sz) arr,arr+sz 47 #define SIZE(v) (int)v.size() 48 #define SORT(v) sort(all(v)) 49 #define REVERSE(v) reverse(ALL(v)) 50 #define SORTA(arr,sz) sort(ALLA(arr,sz)) 51 #define REVERSEA(arr,sz) reverse(ALLA(arr,sz)) 52 #define PERMUTE next_permutation 53 #define TC(t) while(t--) 54 #define forever for(;;) 55 #define PINF 1000000000000 56 #define newline ' ' 57 58 #define test if(1)if(0)cerr 59 using namespace std; 60 using namespace std; 61 typedef vector<int> vi; 62 typedef vector<vi> vvi; 63 typedef pair<int,int> ii; 64 typedef pair<double,double> dd; 65 typedef pair<char,char> cc; 66 typedef vector<ii> vii; 67 typedef long long ll; 68 typedef unsigned long long ull; 69 typedef pair<ll, ll> l4; 70 const double pi = acos(-1.0); 71 72 73 const int MAXNODE = 11000+5; 74 const int SIGMA_SIZE = 26; 75 76 77 int n; 78 string text, input[151]; 79 map<string, int> ms; 80 81 82 inline int idx(char c) { return c-'a'; } 83 int cnt[151]; 84 struct AC 85 { 86 int g[MAXNODE][SIGMA_SIZE], f[MAXNODE], val[MAXNODE], last[MAXNODE]; 87 int sz; 88 int newnode() 89 { 90 memset(g[sz], 0, sizeof(g[sz])); 91 val[sz] = 0; 92 return sz++; 93 } 94 void init() 95 { 96 sz = 0; newnode(); 97 } 98 void insert(const string &str, int v) 99 { 100 int u = 0, n = str.length(); 101 for (int i = 0; i < n; ++i) 102 { 103 int c = idx(str[i]); 104 if (!g[u][c]) g[u][c] = newnode(); 105 u = g[u][c]; 106 } 107 val[u] = v; 108 } 109 void print(int j) 110 { 111 if (j) 112 { 113 ++cnt[val[j]]; 114 115 // operation 116 print(last[j]); 117 } 118 } 119 void find(const string &str) 120 { 121 int n = str.length(), j = 0; 122 for (int i = 0; i < n; ++i) 123 { 124 int c = idx(str[i]); 125 j = g[j][c]; 126 127 if (val[j]) print(j); 128 else if (last[j]) print(last[j]); 129 } 130 } 131 void get_fail() 132 { 133 queue<int> q; f[0] = 0; last[0] = 0; 134 for (int c = 0; c < SIGMA_SIZE; ++c) 135 { 136 int u = g[0][c]; 137 if (u) { f[u] = 0; q.push(u); last[u] = 0; } 138 } 139 while (!q.empty()) 140 { 141 int r = q.front(); q.pop(); 142 for (int c = 0; c < SIGMA_SIZE; ++c) 143 { 144 int u = g[r][c]; 145 if (!u) { g[r][c] = g[f[r]][c]; continue; } 146 q.push(u); 147 int v = f[r]; 148 while (v && !g[v][c]) v = f[v]; 149 f[u] = g[v][c]; 150 last[u] = val[f[u]]?f[u]:last[f[u]]; 151 } 152 } 153 } 154 155 }; 156 157 AC ac; 158 159 int main() 160 { 161 ios::sync_with_stdio(false); 162 cin.tie(0); 163 while (cin >> n && n) 164 { 165 ms.clear(); 166 Reset(cnt, 0); 167 ac.init(); 168 repn(i, n) 169 { 170 cin >> input[i]; 171 ms[input[i]] = i; 172 } 173 cin >> text; 174 for (auto &pr : ms) 175 { 176 ac.insert(pr.first, pr.second); 177 } 178 ac.get_fail(); 179 ac.find(text); 180 int best = 0; 181 for (int i = 1; i <= n; ++i) best = max(best, cnt[i]); 182 cout << best << newline; 183 for (int i = 1; i <= n; ++i) 184 if (cnt[ms[input[i]]] == best) 185 cout << input[i] << newline; 186 } 187 }