题目:戳这里
题意:输出主串中出现的模式串以及该模式串出现的次数。
思路:ac自动机,匹配的时候对应的vis[]+1即可
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<stdio.h> #include<string.h> #include<algorithm> #include<queue> using namespace std; const int maxw = 26; const int maxn = 55; const int time = 1010; const int maxs = 2000005; struct Tire { int tire[maxn*time][maxw]; int fail[maxn*time]; int tag[maxn*time]; int vis[time]; int P, root; int New() { for(int i=0; i<maxw; i++) tire[P][i]=-1; tag[P]=0; return P++; } void init() { P=0; root =New(); } void Insert(char ch[], int id) { int now = root; for(int i=0; ch[i]; i++) { if(tire[now][ch[i]-'A']==-1) tire[now][ch[i]-'A'] = New(); now = tire[now][ch[i]-'A']; } tag[now] = id; } void build() { queue<int>Q; for(int i=0; i<maxw; i++) { if(tire[root][i]!=-1) { fail[tire[root][i]] = root; Q.push(tire[root][i]); } else tire[root][i] = root; } while(!Q.empty()) { int now = Q.front(); Q.pop(); for(int i=0; i<maxw; i++) { if(tire[now][i]!=-1) { fail[tire[now][i]] = tire[fail[now]][i]; Q.push(tire[now][i]); } else tire[now][i] = tire[fail[now]][i]; } } } void match(char ch[]) { int now = root; memset(vis,0,sizeof(vis)); for(int i=0; ch[i]; i++) { if(ch[i]<'A'||ch[i]>'Z') now = root; else { now = tire[now][ch[i]-'A']; int tmp = now; while(tmp!=0) { if(tag[tmp]) vis[tag[tmp]]++; tmp = fail[tmp]; } } } } }; char ch[time][maxn], def[maxs]; int n; Tire ac; int main() { while(~scanf("%d",&n)) { ac.init(); for(int i=1; i<=n; i++) { scanf("%s",ch[i]); ac.Insert(ch[i],i); } ac.build(); scanf("%s",def); ac.match(def); for(int i=1; i<=n; i++) { if(ac.vis[i]!=0) printf("%s: %d ",ch[i],ac.vis[i]); } } }