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HDU 3695-Computer Virus on Planet Pandora(ac自动机)

时间:2015-09-03 23:13:39      阅读:193      评论:0      收藏:0      [点我收藏+]

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题意:

给一个母串和多个模式串,求模式串在母串后翻转后的母串出现次数的的总和。

分析:

模板题

/*#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
using namespace std;
const int maxnode = 250*1000+10000;
const int sigma_size = 26;
struct Trie{
    int ch[maxnode][sigma_size];
    int val[maxnode];     //该单词在模式串中出现的次数
    int last[maxnode];
    int f[maxnode];         //失配数组
    int num[maxnode];     //该单词出现在文本串的次数
    int pre[maxnode];    //该单词的前驱
    int len[maxnode];    //以该单词结尾的单词长度
    int Char[maxnode];     //该单词对应的字母
    int road[maxnode];   //路径压缩优化 针对计算模式串出现的种数
    int sz;
    int Newnode()
    {
        val[sz] = f[sz] = last[sz] = len[sz] = num[sz] = 0;
        memset(ch[sz], 0, sizeof ch[sz]);
        return sz++;
    }
    void init(){
        sz=0;
        Newnode();
    }
    int idx(char c){ return c-‘A‘; }
    int build(char *s){
        int u = 0;
        for(int i = 0, c; s[i] ;i++){
            c = idx(s[i]);
            if(!ch[u][c])
                ch[u][c] = Newnode();
            pre[ch[u][c]] = u;
            Char[ch[u][c]] = s[i];
            len[ch[u][c]] = len[u]+1;
            road[ch[u][c]] = 1;
            u = ch[u][c];
        }
        val[u] = 1;
        num[u] = 0;
        return u;
    }
   void getFail(){
        queue<int> q;
        for(int i = 0; i<sigma_size; i++)
            if(ch[0][i]) q.push(ch[0][i]);
        int r, c, u, v;
        while(!q.empty()){
            r = q.front(); q.pop();
            for(c = 0; c<sigma_size; c++){
                u = ch[r][c];
                if(!u)continue;
                q.push(u);
                v = f[r];
                while(v && ch[v][c] == 0) v = f[v]; //沿失配边走上去 如果失配后有节点 且 其子节点c存在则结束循环
                f[u] = ch[v][c];
            }
        }
    }
    void find(char *T){
    //计算模式串出现的个数:(每种多次出现算多次)
        int j = 0;
        for(int i = 0, c, temp; T[i] ; i++){
            c = idx(T[i]);
            while(j && ch[j][c]==0) j = f[j];
            j = ch[j][c];

            temp = j;
            while(temp){
                num[temp]++;
                temp = f[temp];
            }
        }
    }
    int find_kind(char *T){
    //计算种数, 重复出现的不再计算(若多个询问则要在此处加for(i=0->sz)lu[i]=1;
        int j = 0, i, c, temp,ans=0;
        for(i = 0; T[i]; i++){
            c = idx(T[i]);
            while(j && ch[j][c] == 0) j = f[j];
            j = ch[j][c];
            temp = j;
            while(temp && road[temp]){
                if(val[temp])
                {
                    ++ans;
                    val[temp] = 0;
                }
                road[temp] = 0;
                temp = f[temp];
            }
        }
        return ans;
    }
}ac;
char s[1015], a[5100010], b[5100010], c;
int n, num;
int main() {
    int T; scanf("%d", &T);
    while (T-->0) {
        ac.init();
        scanf("%d", &n);
        while(n--){
            scanf("%s", s);
            ac.build(s);
        }
        ac.getFail();
        int top = 0;
        scanf("%s", a);
        int i = 0,sum=0,id=0;
        while(a[i]!=‘\0‘){
            if(a[i]==‘[‘){
                i++;
                int len=0;
                while(a[i]>=‘0‘&&a[i]<=‘9‘){
                    len=len*10+a[i]-‘0‘;
                    i++;
                }
                for(int j=0;j<len;++j)
                    b[id++]=a[i];
                i+=2;
            }
            else{
                b[id++]=a[i];
                i++;
            }
        }
        b[id]=0;
        //printf("%s\n",b);
        sum+=ac.find_kind(b);
        reverse(b,b+id);
        //printf("%s\n",b);
        sum+=ac.find_kind(b);
        printf("%d\n", sum);
    }
    return 0;
}*/
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <vector>
#include <string>
#include <cctype>
#include <complex>
#include <cassert>
#include <utility>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
typedef pair<int,int> PII;
typedef long long ll;
#define lson l,m,rt<<1
#define pi acos(-1.0)
#define rson m+1,r,rt<<11
#define All 1,N,1
#define read freopen("in.txt", "r", stdin)
#define N 300010
const ll  INFll = 0x3f3f3f3f3f3f3f3fLL;
const int INF= 0x7ffffff;
const int mod =  1000000007;
char P[1010],T[6000000],T1[6000000];
int n;
struct Trie{
    int ch[N][26],val[N],f[N],num;
    void init(){
        num=1;
        memset(ch,0,sizeof(ch));
        memset(val,0,sizeof(val));
        memset(f,0,sizeof(f));
    }
    void build(char *s){
        int u=0,len=strlen(s);
        for(int i=0;i<len;++i)
        {
            int v=s[i]-A;
            if(!ch[u][v]){
                memset(ch[num],0,sizeof(ch[num]));
                ch[u][v]=num++;
            }
            u=ch[u][v];
        }
        val[u]=1;
        //return u;
    }
    void getfail(){
        queue<int>q;
        for(int i=0;i<26;++i)
            if(ch[0][i])
            q.push(ch[0][i]);
        while(!q.empty()){
            int r=q.front();
            q.pop();
            for(int i=0;i<26;++i)
            {
                int u=ch[r][i];
                if(!u){ch[r][i] = ch[f[r]][i];continue;}
                q.push(u);
                int v=f[r];
                while(v&&!ch[v][i])v=f[v];
                f[u]=ch[v][i];
            }
        }
    }
    int find(char *T){
        int u=0,len=strlen(T),total=0;
        for(int i=0;i<len;++i){
            int v=T[i]-A;
            while(u&&ch[u][v]==0)
                u=f[u];
            u=ch[u][v];
            int tmp=u;
            while(tmp){
                if(val[tmp]){
                    total+=val[tmp];
                     val[tmp]=0;
                }
                tmp=f[tmp];
            }
        }
        return total;
    }
}ac;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        ac.init();
        scanf("%d",&n);
        while(n--){
            scanf("%s",P);
            ac.build(P);
        }
        scanf("%s",T);
       // printf("%s",T);
        int id=0,tid=0,i=0;
        char tc;
       while(T[i]!=\0){
            if(T[i]==[){
                i++;
                int len=0;
                while(T[i]>=0&&T[i]<=9){
                    len=len*10+T[i]-0;
                    i++;
                }
                for(int j=0;j<len;++j)
                    T1[id++]=T[i];
                i+=2;
            }
            else{
                T1[id++]=T[i];
                i++;
            }
        }
        T1[id]=0;
       // printf("%s\n",T1);
        int sum=0;
        sum+=ac.find(T1);
        reverse(T1,T1+id);
       //printf("%s\n",T1);
        sum+=ac.find(T1);
    printf("%d\n",sum);
    }
return 0;
}

 

HDU 3695-Computer Virus on Planet Pandora(ac自动机)

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原文地址:http://www.cnblogs.com/zsf123/p/4780838.html

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