2016 ACM/ICPC Asia Regional Qingdao Online 1003（AC自动机）

来源：互联网发布：航拍测绘软件编辑：程序博客网时间：2024/05/16 07:12

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题意：给出一系列的敏感词，如果下面文本以敏感词作为子串，那么输出'*'。

代码：

#include <bits/stdc++.h>using namespace std;const int MAXN = 1000000+100;char str[MAXN];int pos[MAXN];struct Aho_Corasick{    const static int maxnode = 1000000+100;    const static int type = 26;    int next[maxnode][type],fail[maxnode],end[maxnode],Len[maxnode];    int root,L;    int newnode()    {        for(int i = 0;i<type;i++)            next[L][i] = -1;        Len[L]=0;        end[L++] = -1;        return L-1;    }    void init()    {        L = 0;        root = newnode();    }    void insert(char* str)    {        int len = strlen(str);        int now = root;        for(int i = 0;i < len;i++)        {            int id=str[i]-'a';            if(next[now][id] == -1)                next[now][id] = newnode();            now=next[now][id];        }        end[now]=1;//标志结束         Len[now]=len;//记录长度     }    void build()    {        queue<int>Q;        fail[root] = root;        for(int i = 0;i < type;i++)            if(next[root][i] == -1)                next[root][i] = root;            else            {                fail[next[root][i]] = root;                Q.push(next[root][i]);            }        while(!Q.empty())        {            int now = Q.front();            Q.pop();            for(int i = 0;i < type;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]);                }        }    }    void work(char* str)    {        int id;        int len = strlen(str);        int now = root;        memset(pos,0,sizeof(pos));        for(int i = 0;i < len;i++)        {            if(str[i]>='A'&&str[i]<='Z') id=str[i]-'A';            else if(str[i]>='a'&&str[i]<='z') id=str[i]-'a';            else continue;            now = next[now][id];            int temp=now;            while(temp != root)            {                if(end[temp] == 1)                {                    pos[i+1]-=1;                    pos[i-Len[temp]+1]+=1;                    break;                }                temp = fail[temp];            }        }        long long cnt=0;        for(int i=0;i<len;i++)        {            cnt+=pos[i];            if(cnt<=0) printf("%c",str[i]);            else printf("*");        }        printf("\n");    }}AC;int main(){    int T;    int n;    scanf("%d",&T);    while(T--)    {        AC.init();        scanf("%d",&n);        for(int i=1;i<=n;i++)        {            scanf("%s",str);            AC.insert(str);        }        AC.build();        getchar();        gets(str);        AC.work(str);    }    return 0;}

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