Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem B: ASCII Labyrinth

We are trying to construct a labyrinth on a board of sizem × n. Initially, on each square of the board wefind a piece of thin plywood of size 1 × 1 with one of thefollowing three patterns painted on it.

While constructing the labyrinth we may turn the pieces arbitrarilybut each piece must exactly cover a square of the board. We are notallowed to move a piece to another square of the grid.

Given an initial board covered with the pieces, we would like to turnthe pieces in such a way that the patterns on the pieces form at leastone polygonal curve connecting the top left corner square of the boardwith the bottom right square of the board. The picture below presentsan initial state of a board of size 4 × 6 and a labyrinthconstructed from the board in which the above stated goal has beenachieved.

Your task is to read a description of the initial board with thepieces placed on it and to decide whether one can turn the pieces insuch a way that the patterns form a line connecting some edge of thetop left square and some edge of the bottom right square of the board.

The first line of input contains a number c giving the numberof cases that follow. The test data for each case start with twonumbers m and n giving the number of rows and columns onthe board. The remaining lines form an ASCII rendition of the initialboard with the pieces placed on squares. The characters used in therendition are +, -, |, * andspace. See the sample input for the format. The size of the inputboard will be such that m ×  n ≤ 64.

For each case print in a single line how many different paths exist in the solutionsto the labyrinth problem in the format shown below.

Sample input

1
4 6
+---+---+---+---+---+---+
|   |   |   |   |   |   |
|***|***|** |** |** |***|
|   |   | * | * | * |   |
+---+---+---+---+---+---+
|   |   |   |   |   |   |
|   |** |** |***|** |** |
|   | * | * |   | * | * |
+---+---+---+---+---+---+
|   |   |   |   |   |   |
|***|** |***|***|***|** |
|   | * |   |   |   | * |
+---+---+---+---+---+---+
|   |   |   |   |   |   |
|** |   |***|   |** |** |
| * |   |   |   | * | * |
+---+---+---+---+---+---+

Output for sample input

Number of solutions: 2

---

Author: P. Rudnicki

題目輸入只會有三種。
然後左上跟右下也是可以旋轉的。
只要到得了就好。

寫法分兩種狀況討論，水平跟垂直，遇到拐彎的就會從水平便垂直，垂直變水平。

#include <stdio.h>
int n, m, way = 0;
int g[65][65], used[65][65];
void dfs(int x, int y, int d) {// 0 up-down, 1 left-right
    if(x == n-1 && y == m-1) {
        way++;
        return;
    }
    used[x][y] = 1;
    if(d == 0) {
        if(x-1 < 0 || used[x-1][y] || g[x-1][y] == 0) {}
        else {
            if(g[x-1][y] == 1)
                dfs(x-1, y, 0);
            else
                dfs(x-1, y, 1);
        }
        if(x+1 >= n || used[x+1][y] || g[x+1][y] == 0) {}
        else {
            if(g[x+1][y] == 1)
                dfs(x+1, y, 0);
            else
                dfs(x+1, y, 1);
        }
    } else {
        if(y-1 < 0 || used[x][y-1] || g[x][y-1] == 0) {}
        else {
            if(g[x][y-1] == 1)
                dfs(x, y-1, 1);
            else
                dfs(x, y-1, 0);
        }
        if(y+1 >= m || used[x][y+1] || g[x][y+1] == 0) {}
        else {
            if(g[x][y+1] == 1)
                dfs(x, y+1, 1);
            else
                dfs(x, y+1, 0);
        }
    }
    used[x][y] = 0;
}
int main() {
    int testcases;
    int i, j, k;
    char s[4][500];
    scanf("%d", &testcases);
    while(testcases--) {
        scanf("%d %d", &n, &m);
        while(getchar() != '\n');
        while(getchar() != '\n');
        for(i = 0; i < n; i++) {
            for(j = 0; j < 4; j++)
                gets(s[j]);
            for(j = 0, k = 1; j < m; j++, k += 4) {
                if(s[1][k] == ' ')
                    g[i][j] = 0;
                else if(s[1][k+2] == '*') // ***
                    g[i][j] = 1;
                else
                    g[i][j] = 2;
            }
        }
        way = 0;
        dfs(0, 0, 0);
        dfs(0, 0, 1);
        printf("Number of solutions: %d\n", way);
    }
    return 0;
}