CodeForces - [ACM-ICPC Jiaozuo Onsite D]Honeycomb(BFS)

題目鏈接：https://codeforces.com/gym/102028/problem/F
Time limit: 4.0 s Memory limit: 1024 MB

Problem Description

A honeycomb is a mass wax cells built by honey bees, which can be described as a regular tiling of the Euclidean plane, in which three hexagons meet at each internal vertex. The internal angle of a hexagon is $120$ degrees, so three hexagons at a point make a full $360$ degrees. The following figure shows a complete honeycomb with 33 rows and $4$ columns.

Here we guarantee that the first cell in the second column always locates in the bottom right side of the first cell in the first column, as shown above. A general honeycomb may, on the basis of a complete honeycomb, lose some walls between adjacent cells, but the honeycomb is still in a closed form. A possible case looks like the figure below.

Hamilton is a brave bee living in a general honeycomb. Now he wants to move from a starting point to a specified destination. The image below gives a feasible path in a $3×4$ honeycomb from the $1$-st cell in the $2$-nd column to the $1$-st cell in the $4$-th column.

Please help him find the minimum number of cells that a feasible path has to pass through (including the starting point and the destination) from the specified starting point to the destination.

Input

The input contains several test cases, and the first line contains a positive integer $T$ indicating the number of test cases which is up to $10^4$.

For each test case, the first line contains two integers $r$ and $c$ indicating the number of rows and the number of columns of the honeycomb, where $2≤r,c≤10^3$.

The following $(4r+3)$ lines describe the whole given honeycomb, where each line contains at most $(6c+3)$ characters. Odd lines contain grid vertices represented as plus signs ("+") and zero or more horizontal edges, while even lines contain two or more diagonal edges. Specifically, a cell is described as $6$ vertices and at most $6$ edges. Its upper boundary or lower boundary is represented as three consecutive minus signs ("-"). Each one of its diagonal edges, if exists, is a single forward slash ("/") or a single backslash ("\") character. All edge characters will be placed exactly between the corresponding vertices. At the center of the starting cell (resp. the destination), a capital “S” (resp. a capital “T”) as a special character is used to indicate the special cell. All other characters will be space characters. Note that if any input line could contain trailing whitespace, that whitespace will be omitted.

We guarantee that all outermost wall exist so that the given honeycomb is closed, and exactly one “S” and one “T” appear in the given honeycomb. Besides, the sum of $r⋅c$ in all test cases is up to $2×10^6$.

Output

For each test case, output a line containing the minimum number of cells that Hamilton has to visit moving from the starting cell (“S”) to the destination (“T”), including the starting cell and the destination. If no feasible path exists, output -1 instead.

Example

Input

1
3 4

  +---+       +---+
 /     \     /     \
+       +---+       +---+
 \           \     /     \
  +   +   S   +---+   T   +
 /     \     /           /
+       +---+       +   +
 \           \     /     \
  +---+       +---+       +
 /                       /
+       +---+       +   +
 \                 /     \
  +---+       +---+       +
       \     /     \     /
        +---+       +---+

Output

7

Problem solving report:

Description：t組樣例，每組給出若干由字符構成的六邊形單元格，總共n行m列，現在在某兩個單元格中有一個S一個T分別代表起點和重點，問從S到T的最短路徑。
Problem solving：如果圖不是由六邊形給出的，那麼是一個BFS裸題，題目的難點在於如何建圖。我們可以向六邊形的六個方向搜，判斷該方向是否可以通過，如果可以就進入六邊形的中心。就是在搜的方向上和走的步數有所改變，其它的和普通的廣搜沒啥區別。注意，不要用memset，會TLE的！！！

Accepted Code:

/* 
 * @Author: lzyws739307453 
 * @Language: C++ 
 */
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 4e3 + 10, MAXM = 6e3 + 10;
const int dir[6][2] = {{2, 0}, {-2, 0}, {1, 3}, {1, -3}, {-1, 3}, {-1, -3}};
char str[MAXN][MAXM];
bool vis[MAXN][MAXM];
struct edge {
    int x, y, t;
    edge() {}
    edge(int x, int y, int t) : x(x), y(y), t(t) {}
};
int r, c;
bool Judge(int x, int y) {
    if (x >= 0 && x < r && y >= 0 && y < c)
        return true;
    return false;
}
int BFS(int x, int y) {
    queue <edge> Q;
    vis[x][y] = true;
    Q.push(edge(x, y, 1));
    while (!Q.empty()) {
        edge p = Q.front();
        Q.pop();
        if (str[p.x][p.y] == 'T')
            return p.t;
        for (int i = 0; i < 6; i++) {
            int tx = p.x + dir[i][0];
            int ty = p.y + dir[i][1];
            if (str[tx][ty] == ' ' && !vis[tx + dir[i][0]][ty + dir[i][1]] && Judge(tx, ty)) {
                Q.push(edge(tx + dir[i][0], ty + dir[i][1], p.t + 1));
                vis[tx + dir[i][0]][ty + dir[i][1]] = true;
            }
        }
    }
    return -1;
}
int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        int sx, sy;
        scanf("%d%d", &r, &c);
        getchar();
        r = r * 4 + 3;
        c = c * 6 + 3;
        for (int i = 0; i < r; i++) {
            scanf("%[^\n]", str[i]);
            getchar();
            for (int j = 0; str[i][j]; j++) {
                if (str[i][j] == 'S')
                    sx = i, sy = j;
            }
        }
        printf("%d\n", BFS(sx, sy));
        for (int i = 0; i < r; i++)
            for (int j = 0; j < c; j++)
                vis[i][j] = false;
    }
    return 0;
}