poj 1204 Word Puzzles(字典树)

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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;}

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