hdu 3695 Computer Virus on Planet Pandora(AC自动机)

题目连接:hdu 3695 Computer Virus on Planet Pandora

题目大意:给定一些病毒串,要求判断说给定串中包含几个病毒串,包括反转。

解题思路:将给定的字符串展开,然后匹配一次后反转后再匹配一次。

#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 250000;
const int sigma_size = 26;

struct Aho_Corasick {
    int sz, g[maxn][sigma_size];
    int tag[maxn], fail[maxn], last[maxn];

    int vis[300], jump[300];

    void init();
    int idx(char ch);
    void insert(char* str, int k);
    void getFail();
    void match(char* str);
    void put(int u);
    int solve(char* s);
}AC;

int N;
char w[1005], s[5100005];

int main () {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        AC.init();
        scanf("%d", &N);
        for (int i = 1; i <= N; i++) {
            scanf("%s", w);
            AC.insert(w, i);
        }

        int mv = 0, x;
        char ch;
        getchar();
        while(ch = getchar(), ch != ‘\n‘) {
            if (ch == ‘[‘) {
                scanf("%d%c", &x, &ch);
                for (int i = 0; i < x; i++)
                    s[mv++] = ch;
                getchar();
            } else
                s[mv++] = ch;
        }
        s[mv] = ‘\0‘;

        printf("%d\n", AC.solve(s));
    }
    return 0;
}

int Aho_Corasick::solve(char* s) {
    memset(vis, 0, sizeof(vis));

    AC.getFail();
    AC.match(s);
    reverse(s, s + strlen(s));
    AC.match(s);

    int ans = 0;
    for (int i = 1; i <= N; i++)
        if (vis[jump[i]])
            ans++;
    return ans;
}

void Aho_Corasick::init() {
    sz = 1;
    tag[0] = 0;
    memset(g[0], 0, sizeof(g[0]));
}

int Aho_Corasick::idx(char ch) {
    return ch - ‘A‘;
}

void Aho_Corasick::put(int u) {
    if (vis[tag[u]])
        return;

    vis[tag[u]] = 1;

    if (last[u])
        put(last[u]);
}

void Aho_Corasick::insert(char* str, int k) {
    int u = 0, n = strlen(str);

    for (int i = 0; i < n; i++) {
        int v = idx(str[i]);
        if (g[u][v] == 0) {
            tag[sz] = 0;
            memset(g[sz], 0, sizeof(g[sz]));
            g[u][v] = sz++;
        }
        u = g[u][v];
    }
    if (tag[u] == 0)
        tag[u] = k;
    jump[k] = tag[u];
}

void Aho_Corasick::match(char* str) {
    int n = strlen(str), u = 0;
    for (int i = 0; i < n; i++) {
        int v = idx(str[i]);
        while (u && g[u][v] == 0)
            u = fail[u];

        u = g[u][v];

        if (tag[u])
            put(u);
        else if (last[u])
            put(last[u]);
    }
}

void Aho_Corasick::getFail() {
    queue<int> que;

    for (int i  = 0; i < sigma_size; i++) {
        int u = g[0][i];
        if (u) {
            fail[u] = last[u] = 0;
            que.push(u);
        }
    }

    while (!que.empty()) {
        int r = que.front();
        que.pop();

        for (int i = 0; i < sigma_size; i++) {
            int u = g[r][i];

            if (u == 0) {
                g[r][i] = g[fail[r]][i];
                continue;
            }

            que.push(u);
            int v = fail[r];
            while (v && g[v][i] == 0)
                v = fail[v];

            fail[u] = g[v][i];
            last[u] = tag[fail[u]] ? fail[u] : last[fail[u]];
        }
    }
}

郑重声明:本站内容如果来自互联网及其他传播媒体,其版权均属原媒体及文章作者所有。转载目的在于传递更多信息及用于网络分享,并不代表本站赞同其观点和对其真实性负责,也不构成任何其他建议。