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[JSOI 2012] 玄武密码

时间:2018-08-21 23:03:30      阅读:351      评论:0      收藏:0      [点我收藏+]

标签:str   root   continue   tom   tps   uil   oid   class   clu   

[题目链接]

          https://www.lydsy.com/JudgeOnline/problem.php?id=4327

[算法]

        AC自动机
[代码]

        

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e7 + 10;
const int MAXM = 1e5 + 10;
const int MAXLEN = 110;

int n,m;
int ans[MAXM];
char P[MAXN];
char s[MAXM][MAXLEN];

inline int get_value(char a)
{
        if (a == E) return 0;
        if (a == S) return 1;
        if (a == W) return 2;
        if (a == N) return 3;
} 
struct AC_Automation
{
        int tot;
        int root;
        struct Node
        {
                int child[4];
                int fail;
                bool visited;
        } trie[MAXN];
        inline void insert(char *s)
        {
                int now = root;
                int len = strlen(s + 1);    
                for (int i = 1; i <= len; i++)
                {
                        int val = get_value(s[i]);
                        if (!trie[now].child[val]) trie[now].child[val] = ++tot;
                        now = trie[now].child[val];    
                }    
        }        
        inline void rebuild()
        {
                int l,r;
                static int q[MAXN];
                q[l = r = 1] = root;
                trie[root].fail = -1;
                while (l <= r)
                {
                        int cur = q[l];
                        l++;
                        for (int i = 0; i < 4; i++)
                        {
                                if (trie[cur].child[i])
                                {
                                        if (cur == root)
                                                trie[trie[cur].child[i]].fail = 0;
                                        else
                                        {
                                                int p = trie[cur].fail;
                                                while (p != -1)
                                                {
                                                        if (trie[p].child[i]) 
                                                        {
                                                                trie[trie[cur].child[i]].fail = trie[p].child[i];
                                                                break;
                                                        } else p = trie[p].fail;
                                                }
                                        }
                                        q[++r] = trie[cur].child[i];
                                 }
                        }
                }
        }
        inline void getans(char *s)
        {
                int now = root;
                int len = strlen(s + 1);
                for (int i = 1; i <= len; i++)
                {
                        int val = get_value(s[i]);
                        int p = now;
                        while (p != -1)
                        {
                                if (trie[p].child[val]) break;
                                else p = trie[p].fail;
                        }
                        if (p == -1) 
                        {
                                now = 0;
                                continue;
                        } else now = p = trie[p].child[val];
                        while (p)
                        {
                                if (!trie[p].visited)
                                {
                                        trie[p].visited = true;
                                        p = trie[p].fail;
                                } else break;
                        }
                }
        }
        inline void calc(int pos,char *s)
        {
                int now = root;
                int len = strlen(s + 1);
                for (int i = 1; i <= len; i++)
                {
                        int val = get_value(s[i]);
                        now = trie[now].child[val];
                        if (trie[now].visited) ans[pos] = i;
                }
        }
} ACAM;

int main() 
{
        
        scanf("%d%d",&n,&m);
        scanf("%s",P + 1);
        for (int i = 1; i <= m; i++) scanf("%s",s[i] + 1);
        for (int i = 1; i <= m; i++) ACAM.insert(s[i]);
        ACAM.rebuild();
        ACAM.getans(P);
        for (int i = 1; i <= m; i++) ACAM.calc(i,s[i]);
        for (int i = 1; i <= m; i++) printf("%d\n",ans[i]);
        
        return 0;
    
}

 

[JSOI 2012] 玄武密码

标签:str   root   continue   tom   tps   uil   oid   class   clu   

原文地址:https://www.cnblogs.com/evenbao/p/9514949.html

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