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LA 4670 Dominating Patterns (AC自动机)
题意:给定一个一篇文章,然后下面有一些单词,问这些单词在这文章中出现过几次。
析:这是一个AC自动机的裸板,最后在匹配完之后再统计数目就好。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #define debug puts("+++++") //#include <tr1/unordered_map> #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; //using namespace std :: tr1; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const LL LNF = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e6 + 5; const LL mod = 2147493647; const int N = 1e6 + 5; const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1}; const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1}; const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); } inline int gcd(int a, int b){ return b == 0 ? a : gcd(b, a%b); } inline int lcm(int a, int b){ return a * b / gcd(a, b); } int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } const int sigma = 26; const int maxnode = 70 * 150 + 5; struct Aho{ int cnt[155]; int ch[maxnode][sigma]; int f[maxnode]; int val[maxnode]; int last[maxnode]; int sz; void init(){ sz = 1; memset(ch[0], 0, sizeof ch[0]); memset(cnt, 0, sizeof cnt); } int idx(char ch){ return ch - ‘a‘; } void print(int j){ if(j){ ++cnt[val[j]]; print(last[j]); } } void insert(char *s, int v){ int u = 0; while(*s){ int c = idx(*s); if(!ch[u][c]){ memset(ch[sz], 0, sizeof ch[sz]); val[sz] = 0; ch[u][c] = sz++; } u = ch[u][c]; ++s; } val[u] = v; } void find(char *s){ int j = 0; while(*s){ int c = idx(*s); while(j && !ch[j][c]) j = f[j]; j = ch[j][c]; if(val[j]) print(j); else if(last[j]) print(last[j]); ++s; } } void getFail(){ queue<int> q; f[0] = 0; for(int c = 0; c < sigma; ++c){ int u = ch[0][c]; if(u){ f[u] = 0; q.push(u); last[u] = 0; } } while(!q.empty()){ int r = q.front(); q.pop(); for(int c = 0;c < sigma; ++c){ int u = ch[r][c]; if(!u) continue; q.push(u); int v = f[r]; while(v && !ch[v][c]) v = f[v]; f[u] = ch[v][c]; last[u] = val[f[u]] ? f[u] : last[f[u]]; } } } }; Aho aho; char s[155][75]; char t[1000005]; map<string, int> mp; int main(){ while(scanf("%d", &n) == 1 && n){ aho.init(); mp.clear(); for(int i = 1; i <= n; ++i){ scanf("%s", s[i]); aho.insert(s[i], i); mp[s[i]] = i; } scanf("%s", t); aho.getFail(); aho.find(t); int mmax = -1; for(int i = 1; i <= n; ++i) mmax = Max(mmax, aho.cnt[i]); printf("%d\n", mmax); for(int i = 1; i <= n; ++i) if(aho.cnt[mp[s[i]]] == mmax) printf("%s\n", s[i]); } return 0; }
LA 4670 Dominating Patterns (AC自动机)
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