这个题把病毒分为了两种，一种包含可以覆盖，另一种不可以，需要分别求出包含他们的个数，可以把两种都建在一颗tire树上，在最后求得时候判断一下当前节点是属于哪种字符串，如果是不包含的需要判断一下pre[i]+len[i]<=当前位置。
注意会有重复字符串，可以先map处理一下。
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 #include<string> 11 #include<map> 12 using namespace std; 13 #define N 600110 14 #define LL long long 15 #define INF 0xfffffff 16 const double eps = 1e-8; 17 const double pi = acos(-1.0); 18 const double inf = ~0u>>2; 19 const int child_num = 26; 20 char vir[10]; 21 char s[N/6]; 22 int o[N/6],len[N/6],ans[N/6][2],po[N/6]; 23 int cc[N/6]; 24 map<string,int>f; 25 vector<int>ed[N/6]; 26 class AC 27 { 28 private: 29 int ch[N][child_num]; 30 int fail[N]; 31 int Q[N]; 32 int val[N]; 33 int sz; 34 int id[128]; 35 public: 36 void init() 37 { 38 fail[0] = 0; 39 for(int i = 0 ;i < child_num ; i++) 40 id[i+'a'] = i; 41 } 42 void reset() 43 { 44 memset(ch[0],0,sizeof(ch[0])); 45 memset(val,0,sizeof(val)); 46 sz = 1; 47 } 48 void insert(char *a,int key) 49 { 50 int p = 0; 51 for(; *a ; a++) 52 { 53 int d= id[*a]; 54 if(ch[p][d]==0) 55 { 56 memset(ch[sz],0,sizeof(ch[sz])); 57 ch[p][d] = sz++; 58 } 59 p = ch[p][d]; 60 } 61 val[p] = key; 62 } 63 void construct() 64 { 65 int i,head=0,tail = 0; 66 for(i = 0; i < child_num ;i++) 67 { 68 if(ch[0][i]) 69 { 70 fail[ch[0][i]] = 0; 71 Q[tail++] = ch[0][i]; 72 } 73 } 74 while(head!=tail) 75 { 76 int u = Q[head++]; 77 for(i = 0; i < child_num ; i++) 78 { 79 if(ch[u][i]) 80 { 81 fail[ch[u][i]] = ch[fail[u]][i]; 82 Q[tail++] = ch[u][i]; 83 } 84 else ch[u][i] = ch[fail[u]][i]; 85 } 86 } 87 } 88 void work(int m,int kk,int g) 89 { 90 memset(ans,0,sizeof(ans)); 91 memset(po,-1,sizeof(po)); 92 int p = 0,i,k = strlen(s); 93 for(i = 0 ; i < k ; i++) 94 { 95 int d = id[s[i]]; 96 p = ch[p][d]; 97 int tmp = p; 98 while(tmp) 99 { 100 int v = val[tmp]; 101 ans[v][0]++; 102 if(po[v]+len[v]<=i) 103 { 104 po[v] = i; 105 ans[v][1]++; 106 } 107 tmp = fail[tmp]; 108 } 109 } 110 for(i = 1; i <= g ; i++) 111 { 112 for(int j = 0; j < ed[i].size() ; j++) 113 { 114 int v = ed[i][j]; 115 if(o[v]) cc[v] = ans[i][1]; 116 else cc[v] = ans[i][0]; 117 } 118 } 119 printf("Case %d ",kk); 120 for(i = 1 ; i <= m ;i++) 121 printf("%d ",cc[i]); 122 puts(""); 123 } 124 }ac; 125 int main() 126 { 127 int i,m,kk=0; 128 ac.init(); 129 while(scanf("%s",s)!=EOF) 130 { 131 ac.reset(); 132 f.clear(); 133 scanf("%d",&m); 134 int g = 0; 135 for(i = 1 ; i <= m; i++) 136 ed[i].clear(); 137 for(i = 1;i <= m ;i++) 138 { 139 scanf("%d%s",&o[i],vir); 140 if(!f[vir]) 141 { 142 f[vir] = (++g); 143 ac.insert(vir,g); 144 len[g] = strlen(vir); 145 } 146 ed[f[vir]].push_back(i); 147 } 148 ac.construct(); 149 kk++; 150 ac.work(m,kk,g); 151 } 152 return 0; 153 }