LightOJ 1427 Substring Frequency (II) (AC自动机)
Posted dwtfukgv
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题意:给定一个文本串和 n 个子串,问你子串在文本串出现的次数。
析:很明显的AC自动机,只要把先把子串进行失配处理,然后再去用文本串去匹配,在插入子串时就要标记每个串,注意串可能是相同的,这个我错了两次,最后匹配一次就OK了。
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> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e16; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e6 + 10; const int mod = 1e9 + 7; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; 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 bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } char t[maxn]; char s[510][510]; const int maxnode = 510 * 510 + 10; struct Aho{ int ch[maxnode][26]; int f[maxnode]; int val[maxnode]; int last[maxnode]; int cnt[510]; int sz; void init(){ sz = 1; memset(ch[0], 0, sizeof ch[0]); memset(cnt, 0, sizeof cnt); } int idx(char c){ return c - ‘a‘; } 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 getFail(){ queue<int> q; f[0] = 0; for(int c = 0; c < 26; ++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 < 26; ++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]]; } } } 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 print(int j){ if(j){ ++cnt[val[j]]; print(last[j]); } } }; Aho aho; map<string, int> mp; int main(){ int T; cin >> T; for(int kase = 1; kase <= T; ++kase){ scanf("%d", &n); scanf("%s", t); aho.init(); mp.clear(); for(int i = 1; i <= n; ++i){ scanf("%s", s+i); if(mp.count(s[i])) continue; mp[s[i]] = i; aho.insert(s[i], i); } aho.getFail(); aho.find(t); printf("Case %d:\n", kase); for(int i = 1; i <= n; ++i) printf("%d\n", aho.cnt[mp[s[i]]]); } return 0; }
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