zoj3228Searching the String（ac自动机)

链接

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。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
#include<string>
#include<map>
using namespace std;
#define N 600110
#define LL long long
#define INF 0xfffffff
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;
const int child_num = 26;
char vir[10];
char s[N/6];
int o[N/6],len[N/6],ans[N/6][2],po[N/6];
int cc[N/6];
map<string,int>f;
vector<int>ed[N/6];
class AC
{
    private:
    int ch[N][child_num];
    int fail[N];
    int Q[N];
    int val[N];
    int sz;
    int id[128];
    public:
    void init()
    {
        fail[0] = 0;
        for(int i = 0 ;i < child_num ; i++)
        id[i+'a'] = i;
    }
    void reset()
    {
        memset(ch[0],0,sizeof(ch[0]));
        memset(val,0,sizeof(val));
        sz = 1;
    }
    void insert(char *a,int key)
    {
        int p = 0;
        for(; *a ; a++)
        {
            int d= id[*a];
            if(ch[p][d]==0)
            {
                memset(ch[sz],0,sizeof(ch[sz]));
                ch[p][d] = sz++;
            }
            p = ch[p][d];
        }
        val[p] = key;
    }
    void construct()
    {
        int i,head=0,tail = 0;
        for(i = 0; i  < child_num ;i++)
        {
            if(ch[0][i])
            {
                fail[ch[0][i]] = 0;
                Q[tail++] = ch[0][i];
            }
        }
        while(head!=tail)
        {
            int u = Q[head++];
            for(i = 0; i < child_num ; i++)
            {
                if(ch[u][i])
                {
                    fail[ch[u][i]] = ch[fail[u]][i];
                    Q[tail++] = ch[u][i];
                }
                else ch[u][i] = ch[fail[u]][i];
            }
        }
    }
    void work(int m,int kk,int g)
    {
        memset(ans,0,sizeof(ans));
        memset(po,-1,sizeof(po));
        int p = 0,i,k = strlen(s);
        for(i = 0 ; i < k ; i++)
        {
            int d = id[s[i]];
            p = ch[p][d];
            int tmp = p;
            while(tmp)
            {
                int v = val[tmp];
                ans[v][0]++;
                if(po[v]+len[v]<=i)
                {
                    po[v] = i;
                    ans[v][1]++;
                }
                tmp = fail[tmp];
            }
        }
        for(i = 1; i <= g ; i++)
        {
            for(int j = 0; j < ed[i].size() ; j++)
            {
                int v = ed[i][j];
                if(o[v]) cc[v] = ans[i][1];
                else cc[v] = ans[i][0];
            }
        }
        printf("Case %d
",kk);
        for(i = 1 ; i <= m ;i++)
        printf("%d
",cc[i]);
        puts("");
    }
}ac;
int main()
{
    int i,m,kk=0;
    ac.init();
    while(scanf("%s",s)!=EOF)
    {
        ac.reset();
        f.clear();
        scanf("%d",&m);
        int g = 0;
        for(i = 1 ; i <= m; i++)
        ed[i].clear();
        for(i = 1;i <= m ;i++)
        {
            scanf("%d%s",&o[i],vir);
            if(!f[vir])
            {
                f[vir] = (++g);
                ac.insert(vir,g);
                len[g] = strlen(vir);
            }
            ed[f[vir]].push_back(i);
        }
        ac.construct();
        kk++;
        ac.work(m,kk,g);
    }
    return 0;
}