HDU 3695 Computer Virus on Planet Pandora

传送门：http://acm.hdu.edu.cn/showproblem.php?pid=3695

题意：给你n个字串和一个母串，问你有多少个字串与母串匹配（如果某一字串不匹配，那么看它的反转字符串匹不匹配）。

题解：第一道自己做的没看别人题解的AC自动机，而且还是1A，好激动的说，开始向普通AC自动机一样扫一遍，然后记录那些已经被匹配，然后加入未被匹配的反转字符串，再扫一遍，两个的和即位答案。

AC代码：

Accepted

3695

1312MS

30488K

4964 B

G++

XH_Reventon

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <list>
#include <deque>
#include <queue>
#include <iterator>
#include <stack>
#include <map>
#include <set>
#include <algorithm>
#include <cctype>
using namespace std;

#define si1(a) scanf("%d",&a)
#define si2(a,b) scanf("%d%d",&a,&b)
#define sd1(a) scanf("%lf",&a)
#define sd2(a,b) scanf("%lf%lf",&a,&b)
#define ss1(s)  scanf("%s",s)
#define pi1(a)    printf("%d\n",a)
#define pi2(a,b)  printf("%d %d\n",a,b)
#define mset(a,b)   memset(a,b,sizeof(a))
#define forb(i,a,b)   for(int i=a;i<b;i++)
#define ford(i,a,b)   for(int i=a;i<=b;i++)

typedef long long LL;
const int N=33;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
const double eps=1e-8;

#define N 500010
#define L 27   //表示字符串范围
char str[5500010], keyword[1111],ss[5500010];
int head, tail;
char xh[255][1111];
bool vis[255];

struct node
{
    node *fail;
    node *next[L];
    int count;
    int di;
    node() //init内联
    {
        fail = NULL;
        count = 0;
        di=-1;
        for(int i = 0; i < L; ++i)
            next[i] = NULL;
    }
}*q[N];
node *root;

void insert(char *str,int ppp) //建立Trie树
{
    int temp, len;
    node *p = root;
    len = strlen(str);
    for(int i = 0; i < len; ++i)
    {
        temp = str[i] - 'A';  //由题意决定
        if(p->next[temp] == NULL)
            p->next[temp] = new node();
        p = p->next[temp];
    }
    p->count++;
    p->di=ppp;
}


void build_ac() //初始化fail指针，BFS
{
    head= tail = 0;
    q[tail++] = root;
    while(head != tail)
    {
        node *p = q[head++]; //弹出队头
        node *temp = NULL;
        for(int i = 0; i < L; ++i)
        {
            if(p->next[i] != NULL)
            {
                if(p == root) //第一个元素fail必指向根
                    p->next[i]->fail = root;
                else
                {
                    temp = p->fail; //失败指针
                    while(temp != NULL) //2种情况结束：匹配为空or
                    {
                        if(temp->next[i] != NULL) //找到匹配
                        {
                            p->next[i]->fail = temp->next[i];
                            break;
                        }
                        temp = temp->fail;
                    }
                    if(temp == NULL) //为空则从头匹配
                        p->next[i]->fail = root;
                }
                q[tail++] = p->next[i]; //入队
            }
        }
    }
}

int query() //扫描
{
    int index, len, result;
    node *p = root; //Tire入口
    result = 0;
    len = strlen(str);
    for(int i = 0; i < len; ++i)
    {
        index = str[i] - 'A';   //由题意决定
        while(p->next[index] == NULL && p != root) //跳转失败指针
            p = p->fail;
        p = p->next[index];
        if(p == NULL)
            p = root;
        node *temp = p; //p不动，temp计算后缀串
        while(temp != root && temp->count != -1)
        {
            result += temp->count;
            vis[temp->di]=1;
            temp->count = -1;
            temp = temp->fail;
        }
    }
    return result;
}

void deal(node *T)
{//若果有多组数据-多棵树，那么要释放内纯
    int i;
    if(T==NULL)
        return ;
    for(i=0;i<L;i++)
        if(T->next[i]!=NULL)
            deal(T->next[i]);
    free(T);//释放
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(vis,0,sizeof(vis));
        root = new node();
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%s",keyword);
            insert(keyword,i);
            int len=strlen(keyword);
            for(int j=0;j<len;j++)
                xh[i][j]=keyword[len-j-1];
            xh[i][len]='\0';
        }
        build_ac();
        scanf("%s",ss);
        int len=strlen(ss),p=0;
        for(int i=0;i<len;i++)
        {
            if(ss[i]=='[')
            {
                int to=0;
                i++;
                while(ss[i]>='0'&&ss[i]<='9')
                {
                    to=to*10+ss[i]-'0';
                    i++;
                }
                char c=ss[i];
                for(int j=0;j<to;j++)   str[p++]=c;
                i++;
            }
            else
                str[p++]=ss[i];
        }
        str[p]='\0';
        int sum=query();
        deal(root);

        root = new node();
        for(int i=0;i<n;i++)
        {
            if(vis[i]==1)   continue;
            insert(xh[i],i);
        }
        build_ac();
        sum+=query();
        deal(root);
        cout<<sum<<endl;
    }
    return 0;
}
//Accepted	3695	1312MS	30488K	4964 B	G++	XH_Reventon