study from:
https://blog.csdn.net/cindywry/article/details/51919282
https://www.luogu.org/problemnew/show/P3808
1 /* 2 字符出现最多的次数 3 */ 4 #include <cstdio> 5 #include <cstdlib> 6 #include <cmath> 7 #include <cstring> 8 #include <time.h> 9 #include <string> 10 #include <set> 11 #include <map> 12 #include <list> 13 #include <stack> 14 #include <queue> 15 #include <vector> 16 #include <bitset> 17 #include <ext/rope> 18 #include <algorithm> 19 #include <iostream> 20 using namespace std; 21 #define ll long long 22 #define minv 1e-6 23 #define inf 1e9 24 #define pi 3.1415926536 25 #define E 2.7182818284 26 const ll mod=1e9+7;//998244353 27 const int maxn=1e6+10; 28 29 struct node 30 { 31 int c,hav,vis; 32 node* pre; 33 node* next[26]; 34 node* fail; 35 }*tr,*be; 36 queue<node*>q; 37 38 char str[maxn]; 39 40 int main() 41 { 42 node *p,*d,*pos; 43 int n,c,i,j,k,len,num=0,sum=0; 44 tr=(node*) malloc (sizeof(node)); 45 tr->hav=0; 46 tr->vis=0; 47 for (i=0;i<26;i++) 48 tr->next[i]=NULL; 49 scanf("%d",&n); 50 while (n--) 51 { 52 scanf("%s",str); 53 len=strlen(str); 54 pos=tr; 55 for (j=0;j<len;j++) 56 { 57 c=str[j]-97; 58 if (pos->next[c]) 59 pos=pos->next[c]; 60 else 61 { 62 p=(node*) malloc (sizeof(node)); 63 for (k=0;k<26;k++) 64 p->next[k]=NULL; 65 p->c=c; 66 p->hav=0; 67 p->vis=0; 68 pos->next[c]=p; 69 p->pre=pos; 70 pos=p; 71 } 72 } 73 pos->hav++; 74 } 75 be=(node*) malloc (sizeof(node)); 76 for (i=0;i<26;i++) 77 be->next[i]=tr;///all! 78 tr->pre=be; 79 80 q.push(tr); 81 82 while (!q.empty()) 83 { 84 d=q.front(); 85 q.pop(); 86 if (d==tr) 87 d->fail=be; 88 else 89 { 90 pos=d->pre ->fail; 91 c=d->c; 92 while (pos->next[c] == NULL) 93 pos=pos->fail; 94 d->fail=pos->next[c]; 95 } 96 97 for (i=0;i<26;i++) 98 if (d->next[i]) 99 q.push(d->next[i]); 100 } 101 102 be->vis=1; 103 scanf("%s",str); 104 len=strlen(str); 105 pos=tr; 106 for (i=0;i<len;i++) 107 { 108 c=str[i]-97; 109 if (pos->next[c]) 110 pos=pos->next[c]; 111 else 112 { 113 while (pos->next[c]==NULL) 114 pos=pos->fail; 115 pos=pos->next[c]; 116 } 117 d=pos; 118 while (d->vis==0) 119 { 120 d->vis=1; 121 sum+=d->hav; 122 d=d->fail; 123 } 124 } 125 cout<<sum; 126 return 0; 127 } 128 /* 129 2 130 ab 131 ac 132 abc 133 134 2 135 ababc 136 abc 137 ababc 138 139 3 140 ab 141 abc 142 abca 143 ababc 144 145 6 146 ab 147 abc 148 abc 149 abc 150 abab 151 ababc 152 ababc 153 */