学步园

题目链接：poj 1204 Word Puzzles

题目大意：给定一个有字符组成的N行M列的矩阵，就这是Q次查询，每次查询包括一个字符串，要求在矩阵中找到起

始点以及方向。

解题思路：对查询建立字典树，然后暴力枚举矩阵中的起点和方向。数据有点弱，就这样给过了。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 1e6+5;
const int maxc = 1005;
const int sigma_size = 26;
const int dir[8][2] = {{-1, 0}, {-1, 1}, {0, 1}, {1, 1}, {1, 0}, {1, -1}, {0, -1}, {-1, -1}};

struct Tire {
    int sz, C;
    int g[maxn][sigma_size];
    int val[maxn];

    void init();
    int idx(char ch);
    void insert(char* s, int x);
    void find(int x, int y, int d);
}T;

int N, M, Q, X[maxc], Y[maxc], D[maxc];
char G[maxc][maxc], W[maxc];

void solve () {
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
            for (int d = 0; d < 8; d++) {
                T.find(i, j, d);
                if (T.C == 0)
                    return;
            }
        }
    }
}

int main () {
    while (scanf("%d%d%d", &N, &M, &Q) == 3) {
        for (int i = 0; i < N; i++)
            scanf("%s", G[i]);

        T.init();
        for (int i = 1; i <= Q; i++) {
            scanf("%s", W);
            T.insert(W, i);
        }

        solve();
        for (int i = 1; i <= Q; i++)
            printf("%d %d %c\n", X[i], Y[i], D[i] + 'A');
    }
    return 0;
}

void Tire::init() {
    C = 0;
    sz = 1;
    val[0] = 0;
    memset(g[0], 0, sizeof(g[0]));
}

int Tire::idx (char ch) {
    return ch - 'A';
}

void Tire::find(int x, int y, int d) {
    int p = x, q = y, u = 0;

    for (int i = 1; i; i++) {
        if (p < 0 || p >= N || q < 0 || q >= M)
            break;
        int v = idx(G[p][q]);

        if (g[u][v] == 0)
            break;

        u = g[u][v];
        p += dir[d][0];
        q += dir[d][1];

        if (val[u]) {
            X[val[u]] = x;
            Y[val[u]] = y;
            D[val[u]] = d;
            C--;
            val[u] = 0;
        }
    }
}

void Tire::insert(char* s, int x) {
    int u = 0, n = strlen(s);

    for (int i = 0; i < n; i++) {
        int v = idx(s[i]);

        if (g[u][v] == 0) {
            val[sz] = 0;
            memset(g[sz], 0, sizeof(g[sz]));
            g[u][v] = sz++;
        }
        u = g[u][v];
    }
    C++;
    val[u] = x;
}