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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10324 Accepted Submission(s): 3633
/* hdu 3065 AC自动机(各子串出现的次数) 给你m个子串,然后从一个字符串中查找这些子串哪些出现过,出现了多少次 在用ed对子串进行标记。 查询过程中遇到ed则对在相应的子串上 +1 hhh-2016-04-23 21:08:09 */ #include <iostream> #include <vector> #include <cstring> #include <string> #include <cstdio> #include <queue> #include <algorithm> #include <functional> #include <map> using namespace std; #define lson (i<<1) #define rson ((i<<1)|1) typedef long long ll; typedef unsigned int ul; const int maxn = 40010; const int mod = 10007; int ans[1005]; int tot; struct Tire { int nex[1000*55][130],fail[1000*55],ed[1000*55]; int root,L; int newnode() { for(int i = 0; i < 130; i++) nex[L][i] = -1; ed[L++] = 0; return L-1; } void ini() { L = 0,root = newnode(); } void inser(char buf[],int id) { int len = strlen(buf); int now = root; for(int i = 0; i < len; i++) { int ta = buf[i]; if(nex[now][ta] == -1) nex[now][ta] = newnode(); now = nex[now][ta]; } ed[now] = id; } void build() { queue<int >q; fail[root] = root; for(int i = 0; i < 130; i++) if(nex[root][i] == -1) nex[root][i] = root; else { fail[nex[root][i]] = root; q.push(nex[root][i]); } while(!q.empty()) { int now = q.front(); q.pop(); for(int i = 0; i < 130; i++) { if(nex[now][i] == -1) nex[now][i] = nex[fail[now]][i]; else { fail[nex[now][i]] = nex[fail[now]][i]; q.push(nex[now][i]); } } } } void query(char buf[]) { tot = 0; int cur = root; int len = strlen(buf); for(int i = 0; i < len; i++) { cur = nex[cur][(int)buf[i]]; int tp = cur; while(tp != root) { if(ed[tp]) ans[ed[tp]]++; tp = fail[tp]; } } } }; Tire ac; char buf[1005][55]; char to[2000100]; int main() { int n; while(scanf("%d",&n) != EOF) { ac.ini(); for(int i = 1; i <= n; i++) { scanf("%s",buf[i]); ac.inser(buf[i],i); } ac.build(); memset(ans,0,sizeof(ans)); scanf("%s",to); ac.query(to); for(int i = 1; i <= n; i++) { if(ans[i] > 0) { printf("%s: %d\n",buf[i],ans[i]); } } } return 0; }
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原文地址:http://www.cnblogs.com/Przz/p/5449297.html