hdu5384

来源：互联网发布：电力监控软件编辑：程序博客网时间：2024/04/29 20:26
题意：给你n个母串，m个匹配串，让你求出对于每个母串所有匹配串出现的次数和。
思路：完完全全邝斌的模板啊。。。注释掉一行代码就能a。。。。
代码：
#include <algorithm>#include <iostream>#include <sstream>#include <cstdlib>#include <cstring>#include <iomanip>#include <cstdio>#include <string>#include <bitset>#include <vector>#include <queue>#include <stack>#include <cmath>#include <list>#include <map>#include <set>#define sss(a,b,c) scanf("%d%d%d",&a,&b,&c)#define mem1(a) memset(a,-1,sizeof(a))#define mem(a) memset(a,0,sizeof(a))#define ss(a,b) scanf("%d%d",&a,&b)#define s(a) scanf("%d",&a)#define p(a) printf("%d\n", a)#define INF 0x3f3f3f3f#define w(a) while(a)#define PI acos(-1.0)#define LL long long#define eps 10E-9#define N 100010#define mod 258280327const int SIGMA_SIZE=26;const int MAXN=100010;const int MAXNODE=600010;using namespace std;void mys(int& res){    int flag=0;    char ch;    while(!(((ch=getchar())>='0'&&ch<='9')||ch=='-'))        if(ch==EOF)  res=INF;    if(ch=='-')  flag=1;    else if(ch>='0'&&ch<='9')  res=ch-'0';    while((ch=getchar())>='0'&&ch<='9')  res=res*10+ch-'0';    res=flag?-res:res;}void myp(int a){    if(a>9)        myp(a/10);    putchar(a%10+'0');}/*************************THE END OF TEMPLATE************************/char str1[N][10001];char s[N];int m, ans;struct ac_trie{        int next[500010][26], ed[500010], fail[500010];        int root, L;        int newnode(){            for(int i=0; i<26; i++) next[L][i] = -1;            ed[L++] = 0;            return L-1;        }        void init(){            L = 0;            root = newnode();        }        void m_insert(char *s){            int len = strlen(s);            int now = root;            for(int i = 0; i<len; i++){                if(next[now][s[i]-'a'] == -1) next[now][s[i]-'a'] = newnode();                now = next[now][s[i]-'a'];            }            ed[now]++;        }        void getfail(){            queue<int>Q;            fail[root] = root;            for(int i=0; i<26; i++)                if(next[root][i] == -1) next[root][i] = root;                else{                    fail[next[root][i]] = root;                    Q.push(next[root][i]);                }            w(!Q.empty()){                int now = Q.front();                Q.pop();                for(int i=0; i<26; i++){                    if(next[now][i] == -1) next[now][i] = next[fail[now]][i];                    else{                        fail[next[now][i]] = next[fail[now]][i];                        Q.push(next[now][i]);                    }                }            }        }        int query(char *s){            int len = strlen(s);            int now = root;            int res = 0;            for(int i=0; i<len; i++){                now = next[now][s[i]-'a'];                int temp = now;                w(temp!=root){                   res += ed[temp];                   //ed[temp] = 0;                   temp = fail[temp];                }            }            return res;        }};ac_trie ac;int main(){    int t, n;    s(t);    w(t--){         ss(n, m);         ac.init();         for(int i=0; i<n; i++) scanf("%s",str1[i]);         for(int i=0; i<m; i++){            scanf("%s", s);            ac.m_insert(s);         }         ac.getfail();         for(int i=0; i<n; i++)  p(ac.query(str1[i]));    }    return 0;}
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