hdu 1277 全文检索 (直接映射查找 || 自动机)

Problem - 1277

　　无聊做水题的时候发现的一道题目。这道题第一反应可以用自动机来解决。当然，条件是各种限制，从而导致可以用直接映射标记的方法来搜索。具体的做法就像RK算法一样，将字符串hash成一个数，这里每一个关键字前四位都是不同的，这样就有利于hash搜索了。当前四位匹配的时候，就可以搜索整个串是否完全匹配。这整个的复杂度大概是O(n*m)，n是全文测长度，m是关键字的长度。

自动机代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <queue>
#include <set>

using namespace std;

const int kind = 10;
struct Node {
    Node *next[kind];
    Node *fail;
    string label;
    Node() {
        memset(next, 0, sizeof(next));
        fail = NULL;
        label = "";
    }
    Node(char *tmp) {
        memset(next, 0, sizeof(next));
        fail = NULL;
        label = tmp;
    }
} ;

struct Trie {
    Node *Root;
    Trie() { Root = new Node();}

    void insert(char *s, char *label) {
        Node *p = Root;
        while (*s) {
            int idx = *s - '0';
            if (!p->next[idx]) {
                p->next[idx] = new Node();
                p->next[idx]->fail = p;
            }
            p = p->next[idx];
            s++;
        }
        p->label = label;
    }
    void build() {
        queue<Node *> Q;
        while (!Q.empty()) Q.pop();
        Q.push(Root);
        Root->fail = NULL;
        Node *cur, *p;
        while (!Q.empty()) {
            cur = Q.front();
            Q.pop();
            for (int i = 0; i < kind; i++) {
//                cout << cur->next[i] << ' ';
                if (cur->next[i]) {
//                    cout << i << ' ';
                    Q.push(cur->next[i]);
                    p = cur->fail;
                    while (p) {
                        if (p->next[i]) {
                            cur->next[i]->fail = p->next[i];
                            break;
                        }
                        p = p->fail;
                    }
                    if (!p) cur->next[i]->fail = Root;
                }
            }
        }
    }
    void query(char *s) {
        set<string> vis;
        bool pnt = false;
        Node *cur = Root;
        while (*s) {
            int idx = *(s++) - '0';
            while (cur && !cur->next[idx]) cur = cur->fail;
            cur = cur ? cur->next[idx] : Root;
            if (vis.find(cur->label) != vis.end()) continue;
            if (cur->label.length()) {
                if (!pnt) {
                    printf("Found key:");
                    pnt = true;
                }
                cout << ' ' << cur->label;
                vis.insert(cur->label);
            }
        }
        if (!pnt) puts("No key can be found !");
        cout << endl;
    }
} trie;

char str[66666], buf[111];

int main() {
//    freopen("in", "r", stdin);
    int n, m;
    while (cin >> n >> m) {
        trie = Trie();
        gets(buf);
        str[0] = 0;
        while (n-- && gets(buf)) strcat(str, buf);
        gets(buf);
        while (m-- && gets(buf)) {
            char *p = buf;
            while (*(p++) != ']') ;
            *(p++) = 0;
            while (*p == ' ') p++;
            trie.insert(p, buf);
//            cout << p << ' ' << buf << " inserted!"<< endl;
        }
        trie.build();
//        cout << "built!!" << endl;
        trie.query(str);
    }
    return 0;
}

UPD：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>

using namespace std;

const int HASHLEN = 5;
const int M = 111111;
const int N = 11111;
string label[N], key[N];
bool vis[N];
int id[M];
char str[6 * N], buf[N];

bool check(int t, char *p) {
    int len = key[t].length();
    for (int i = 0; i < len; i++) {
        if (key[t][i] != p[i]) return false;
    }
    return true;
}

void work(char *p) {
    bool pnt = false;
    char *q = p;
    int cur = 0, ep = 10;
    for (int i = 0; i < HASHLEN - 1; i++) cur = cur * 10 + *(q++) - '0', ep *= 10;
    while (*q && *p) {
//        cout << cur << endl;
        cur = cur * 10 + *q - '0';
        cur %= ep;
        int t = id[cur];
        if (t != -1 && !vis[t]) {
            if (check(t, p)) {
                if (!pnt) printf("Found key:");
                pnt = vis[t] = true;
                cout << ' ' << label[t];
            }
        }
        p++, q++;
    }
    if (!pnt) printf("No key can be found !");
    cout << endl;
}

int main() {
//    freopen("in", "r", stdin);
    int n, m;
    while (cin >> n >> m) {
        memset(id, -1, sizeof(id));
        memset(vis, 0, sizeof(vis));
        gets(buf);
        str[0] = 0;
        while (n-- && gets(buf)) strcat(str, buf);
        gets(buf);
        for (int i = 0; i < m; i++) {
            gets(buf);
            char *p = buf;
            while (*p && *p != ']') p++;
            *(++p) = 0;
            p++;
            while (*p && *p == ' ') p++;
            label[i] = buf;
            key[i] = p;
//            cout << label[i] << ' ' << key[i] << endl;
            int cid = 0;
            for (int i = 0; i < HASHLEN; i++) cid = cid * 10 + p[i] - '0';
//            cout << cid << endl;
            if (id[cid] != -1) { puts("shit!"); while (1) ;}
            id[cid] = i;
        }
        work(str);
    }
    return 0;
}

　　不知道为什么这个代码能过G++但是C++过不了，应该是代码里面有bug。另外，数据似乎跟描述的有点出入，说好的不同key值的前四位不会相同，可是写了句来试它的数据时就发现被骗了，所以后来用了5位，过了！

——written by Lyon