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

Universidade da Beira Interior 21-10-2006 

Problem G

You have been blindfolded and brought to a strange complex of mazes. Each maze is divided in squares, each one with a strange circular plate with arrows on it. The start of each maze is on the upper left corner and the exit is always on the bottom right. The following figure shows an example maze:

During a single time unit you can go from a square to one of the four adjacent squares. But you can only follow the directions that the arrows on your current square point. Trying to do other thing will get you killed. Each time you pass from one square to another, all plates rotate 90 degrees in clockwise direction, changing the way the maze looks. Is it possible to get to the exit? And what is the best and fastest way to do that?

Problem

Given a particular maze in the conditions described above, your task is to discover how much time does it take the quickest path from the upper left corner to the bottom right corner (the exit). You must also discover if that is not possible.

Input

The input file contains several test cases, each of them as describes below. The input will start with a single line containing two numbers separated by a single space: R and C indicating respectively the number of rows and columns of the maze (2   R,C   500).

Then there are exactly (R*C)-1 lines indicating in which directions are the arrows of each square pointing. These lines are given in a specific order, starting from the north to the south, and from the west to the east. This is, if we use the notation (row,column), the lines are given in the order (1,1),(1,2),...,(1,C),(2,1),...,(2,N),...,(R,1),...,(R,C-1). The bottom right corner (R,C) is not given, since it is always the exit.

Each of these lines contains a single string of length one to four chars, indicating the arrows of the plate on time 0. The chars belong to the set N,S,W,E and represent respectively an arrow pointing to North, South, West and East. See example input 1 for a representation of the example maze given in figure 1. There will not be repeated chars on the same line and the chars can appear in any order.

Output

For each test case, the output should contain a single line with an integer that represents the time taken by the quickest path from the start (always square (1,1)) to the exit (always (R,C)). Remember that the plates always rotate when you change your current square (therefore is does not help to stay on the same place waiting for a rotation - it won't happen!) and you can only follow the directions that the arrows point on the present time you are on that square.

If there is no path from the start to the exit you should print "no path to exit".

Sample Input

2 2
NES
S
WS
3 2
NSWE
SW
SEW
NEW
SN

Sample Output

4
no path to exit

Maratona Inter-Universitária de Programação 2006 MIUP'2006 

---

Author: Pedro Ribeiro (Universidade do Porto)#include <stdio.h>

題目描述：

每一格都有可轉移到其它相鄰的格子的限制，每走一步所有限制順時針轉 90 度，

求走到右下角 Exit 的最少時間。

題目解法：

很容易想，每一格最多 4 種狀態，也就是當前轉了 90, 180, 270, 360 度。

藉此進行 bfs 計算即可。

#include <string.h>
#include <queue>
#include <algorithm>
using namespace std;
int n, m;
int dx[] = {-1,0,1,0};//NESW
int dy[] = {0,1,0,-1};
int g[505][505][4];
int dist[505][505][4];
int trans(char c) {
    switch(c) {
        case 'N':return 0;
        case 'E':return 1;
        case 'S':return 2;
        case 'W':return 3;
    }
    return -1;
}
void bfs() {
    queue<int> X, Y, D;
    int i, j, k;
    int x, y, d;
    int tx, ty;
    for(i = 0; i < n; i++)
        for(j = 0; j < m; j++)
            dist[i][j][0] = dist[i][j][1] = dist[i][j][2] = dist[i][j][3] = 0;
    X.push(0), Y.push(0), D.push(0);
    while(!X.empty()) {
        x = X.front(), X.pop();
        y = Y.front(), Y.pop();
        d = D.front(), D.pop();
        for(i = 0; i < 4; i++) {
            if(g[x][y][i] == 0) continue;
            tx = x+dx[(i+d)%4], ty = y+dy[(i+d)%4];
            if(tx < 0 || ty < 0 || tx >= n || ty >= m)
                continue;
            if(dist[tx][ty][(d+1)%4] == 0) {
                dist[tx][ty][(d+1)%4] = dist[x][y][d]+1;
                X.push(tx), Y.push(ty), D.push((d+1)%4);
            }
        }
    }
    int ret = 0xfffffff;
    for(i = 0; i < 4; i++)
        if(dist[n-1][m-1][i])
            ret = min(ret, dist[n-1][m-1][i]);
    if(ret == 0xfffffff)
        puts("no path to exit");
    else
        printf("%d\n", ret);
}
int main() {
    int i, j, k;
    char s[10];
    while(scanf("%d %d", &n, &m) == 2) {
        while(getchar() != '\n');
        memset(g, 0, sizeof(g));
        for(i = 0; i < n; i++) {
            for(j = 0; j < m; j++) {
                if(i == n-1 && j == m-1)
                    continue;
                gets(s);
                for(k = 0; s[k]; k++)
                    g[i][j][trans(s[k])] = 1;
            }
        }
        bfs();
    }
    return 0;
}