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id=1719">点击打开链接

题意：真正的题意是，告诉你一些字符串。然后告诉你非常多个字符格子，问这些字符串是否能在字符格子中连起来，在格子中对角线也觉得是连在一起的。假设格子中的字符是q，事实上是代表着qu

思路：这题迷之英语。各种猜题意啊，，只是运气好比較早就猜中了。嘿嘿嘿

懂题意了后就非常easy了，DFS各种搜即可了，由于数据范围比較小

#include <map>

#include <set>

#include <cmath>

#include <ctime>

#include <stack>

#include <queue>

#include <cstdio>

#include <cctype>

#include <bitset>

#include <string>

#include <vector>

#include <cstring>

#include <iostream>

#include <algorithm>

#include <functional>

#define fuck(x) cout << "[" << x << "]"

#define FIN freopen("input.txt", "r", stdin)

#define FOUT freopen("output.txt", "w+", stdout)

using namespace std;

typedef long long LL;

typedef pair<int, int> PII;

const int MX = 200 + 5;

string A[MX];

int n, m;

char stemp[MX];

char S[10][10], vis[10][10];

bool DFS(int id, int i, int x, int y, char w) {

    int len = A[id].length();

    if(A[id][i] != w) return false;

    if(w == 'q') {

        if(!(i + 1 < len && A[id][i+1] == 'u')) return false;

        else i++;

    }

    if(i == len - 1) return true;

    vis[x][y] = 1;

    for(int dx = -1; dx <= 1; dx++) {

        for(int dy = -1; dy <= 1; dy++) {

            if(dx == 0 && dy == 0) continue;

            int nx = dx + x, ny = dy + y;

            if(nx < 0 || nx > m || ny < 0 || ny > m || vis[nx][ny]) continue;

            if(DFS(id, i + 1, nx, ny, S[nx][ny])) {

                vis[x][y] = 0;

                return true;

            }

        }

    }

    vis[x][y] = 0;

    return false;

}

int main() {

    //FIN;

    while(~scanf("%d", &n)) {

        for(int i = 1; i <= n; i++) {

            scanf("%s", stemp);

            A[i] = string(stemp);

        }

        sort(A + 1, A + 1 + n);

        while(scanf("%d", &m), m) {

            for(int i = 1; i <= m; i++) {

                scanf("%s", S[i] + 1);

            }

            for(int id = 1; id <= n; id++) {

                bool sign = false;

                for(int i = 1; i <= m; i++) {

                    for(int j = 1; j <= m; j++) {

                        if(DFS(id, 0, i, j, S[i][j])) {

                            sign = true; break;

                        }

                    }

                    if(sign) break;

                }

                if(sign) printf("%s\n", A[id].c_str());

            }

            printf("-\n");

        }

    }

    return 0;

}