http://acm.hdu.edu.cn/showproblem.php?pid=2896

题意:中文......

思路:

这里就知识循环枚举每一个网址即可,set记录每个网址包含的病毒的编号。

//#pragma comment(linker,"/STACK:327680000,327680000")
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define ll long long
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define lowbit(x)   (x)&(-x)
#define Read()  freopen("din.txt", "r", stdin)
#define Write() freopen("dout.txt", "w", stdout);


#define M 57
#define N 1000007
using namespace std;

struct node
{
    int cnt;
    node *fail;
    node *next[95];
    void newnode()
    {
        cnt = 0;
        fail = NULL;
        for (int i = 0; i < 95; ++i)
        {
            next[i] = NULL;
        }
    }
};

set<int>iSet[1007];

struct AC_Automat
{
public:

    node *q[N],*root,H[N];
    int fr,tl;
    int t;

    void init()
    {
        fr = tl = 0;
        t = 0;
        H[t].newnode();
        root = &H[t++];
        for (int i = 0; i <= 1000; ++i) iSet[i].clear();
    }
    void insert(char *s,int no)
    {
        int i,k;
        int len = strlen(s);
        node *p = root;
        for (i = 0; i < len; ++i)
        {
            k = s[i] - 32;
            if (p->next[k] == NULL)
            {
                H[t].newnode();
                p->next[k] = &H[t++];
            }
            p = p->next[k];
        }
        p->cnt = no;
    }
    void build()
    {
        int i;
        root->fail = NULL;
        q[tl] = root;
        node *p;
        while (fr <= tl)
        {
            node *tmp = q[fr++];
            for (i = 0; i < 95; ++i)
            {
                if (tmp->next[i])
                {
                    if (tmp == root) tmp->next[i]->fail = root;
                    else
                    {
                        p = tmp->fail;
                        while (p != NULL)
                        {
                            if (p->next[i])
                            {
                                tmp->next[i]->fail = p->next[i];
                                break;
                            }
                            p = p->fail;
                        }
                        if (p == NULL) tmp->next[i]->fail = root;
                    }
                    q[++tl] = tmp->next[i];
                }
            }
        }
    }
    void query(char *s,int no)
    {
        int i,k;
        int len = strlen(s);
        node *p = root;
        for (i = 0; i < len; ++i)
        {
            k = s[i] - 32;
            while (p->next[k] == NULL && p != root)
            p = p->fail;

            p = p->next[k];
            if (p == NULL) p = root;

            node *tmp = p;
            while (tmp != root)
            {
                if (tmp->cnt != 0)
                {
                    iSet[no].insert(tmp->cnt);
                }
                tmp = tmp->fail;
            }
        }
    }
}ac;

int n,m;
char s1[202],s2[10007];

int main()
{
    int i;
    while (~scanf("%d",&n))
    {
        ac.init();
        for (i = 1; i <= n; ++i)
        {
            scanf("%s",s1);
            ac.insert(s1,i);
        }
        ac.build();
        scanf("%d",&m);
        for (i = 1; i <= m; ++i)
        {
            scanf("%s",s2);
            ac.query(s2,i);
        }
        int tol = 0;
        set<int>::iterator it;
        for (i = 1; i <= m; ++i)
        {
            if (iSet[i].size() >= 1)
            {
                printf("web %d:",i);
                tol++;
                for (it = iSet[i].begin(); it != iSet[i].end(); ++it)
                {
                    printf(" %d",*it);
                }
                 printf("\n");
            }

        }
        printf("total: %d\n",tol);
    }
    return 0;
}