HDU 3065 病毒侵襲持續中（AC自動機）

2021-07-10 16:29:00 字數 1521 閱讀 4169

題意：求n個串在乙個很長的串中出現的次數。

思路： 典型的ac自動機模板題。

細節參見**：

#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#define max(a,b) ((a)>(b)?(a):(b))


#define min(a,b) ((a)<(b)?(a):(b))


using namespace std;


typedef long long ll;


typedef long double ld;


const ld eps = 1e-9, pi = 3.1415926535897932384626433832795;


const int mod = 1000000000 + 7;


const int inf = 0x3f3f3f3f;


const int maxs = 1000000*2 + 10;


// & 0x7fffffff


const int seed = 131;


const int sigma_size = 199;


const ll inf64 = ll(1e18);


const int maxn = 100;


const int maxnode = 1000*55 + 10;


int t,n,m;


int ch[maxnode][sigma_size];


int f[maxnode]; // fail函式


int val[maxnode]; // 每個字串的結尾結點都有乙個非0的val


int last[maxnode]; // 輸出鍊錶的下乙個結點


int cnt[maxs];


int sz;


mapms;


void init() 


// 字元c的編號


int idx(char c) 


// 插入字串。v必須非0


void _insert(char *s, int v) 


u = ch[u][c];


}val[u] = v;


// ms[string(s)] = v;


} // 遞迴列印以結點j結尾的所有字串


void print(int j) 


}// 在t中找模板


void _find(char* t) 


}// 按bfs順序計算fail


while(!q.empty()) 


}}char buf[1006][55];


char s[2000000 + 10];


int main() 


getfail();


scanf("%s",s);


_find(s);


for(int i=1;i<=n;i++) 


}return 0;


}

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