[POJ] 3083.Children of the Candy Corn

原题传送门
思路：
输出三种结果： 摸左边墙走，摸右边墙走，最短路线

注意点：
方向是按自己当前方向来算的，第一个路线的方向循环是左上右下，第二个路线的循环是右上左下。

以左路线为例：
思考得之，进入方格的方向是上一格前进时朝向的方向，而当前应该前进的方向是循环内的上一方向。即：如果此时进入方向N，然后应该先向W前进。

DFS运算前两个，BFS运算最短路径

#include <algorithm>
#include <bitset>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>

using namespace std;
/***************************************/
#define ll long long
#define int64 __int64
/***************************************/
const int INF = 0x7f7f7f7f;
int w, h;
int leftMoveY[4] = {-1, 0, 1, 0};
int leftMoveX[4] = {0, -1, 0, 1}; //左上右下
int rightMoveY[4] = {1, 0, -1, 0};
int rightMoveX[4] = {0, -1, 0, 1}; //右上左下
char chess[50][50];
int vis[50][50];
int leftStep, rightStep, shortStep;
int leftStepTemp, rightStepTemp;
int shortStepTemp[50][50];

void leftDFS(int x, int y, int d)
{
  if (chess[x][y] == 'E')
  {
    leftStep = leftStepTemp;
    return;
  }
  int x1, y1;
  if (d == 0)
    d = 3;
  else
    d--;
  for (int i = d; i < 4; i++)
  {
    x1 = x + leftMoveX[i];
    y1 = y + leftMoveY[i];
    if (x1 > 0 && x1 <= h && y1 > 0 && y1 <= w && chess[x1][y1] != '#')
    {
      leftStepTemp++;
      leftDFS(x1, y1, i);
    }
    if (leftStep)
      return;
    if (i == 3)
      i = -1;
  }
}
void rightDFS(int x, int y, int d)
{
  if (chess[x][y] == 'E')
  {
    rightStep = rightStepTemp;
    return;
  }
  int x1, y1;
  if (d == 0)
    d = 3;
  else
    d--;
  for (int i = d; i < 4; i++)
  {
    x1 = x + rightMoveX[i];
    y1 = y + rightMoveY[i];
    if (x1 > 0 && x1 <= h && y1 > 0 && y1 <= w && chess[x1][y1] != '#')
    {
      rightStepTemp++;
      rightDFS(x1, y1, i);
    }
    if (rightStep)
      return;
    if (i == 3)
      i = -1;
  }
}

void BFS(int x, int y)
{
  queue<int> Q;
  Q.push(x);
  Q.push(y);
  int x1, y1;

  while (!Q.empty())
  {
    x1 = Q.front();
    Q.pop();
    y1 = Q.front();
    Q.pop();

    if (vis[x1][y1] == 1)
      continue;

    vis[x1][y1] = 1;
    for (int i = 0; i < 4; i++)
    {
      int x2 = x1 + leftMoveX[i];
      int y2 = y1 + leftMoveY[i];

      if (!vis[x2][y2] && ((chess[x2][y2] == '.') || (chess[x2][y2] == 'E')))
      {
        Q.push(x2);
        Q.push(y2);
        shortStepTemp[x2][y2] = shortStepTemp[x1][y1] + 1;

        if (chess[x2][y2] == 'E')
        {
          shortStep = shortStepTemp[x2][y2];
          return;
        }
      }
    }
  }
}

int main()
{
  int n;
  cin >> n;
  while (n--)
  {
    memset(chess, 0, sizeof(chess));
    memset(vis, 0, sizeof(vis));
    memset(shortStepTemp, 0, sizeof(shortStepTemp));
    leftStep = rightStep = shortStep = leftStepTemp = rightStepTemp = 0;
    cin >> w >> h;
    int sx, sy;
    for (int i = 1; i <= h; i++)
      for (int j = 1; j <= w; j++)
      {
        cin >> chess[i][j];
        if (chess[i][j] == 'S')
          sx = i, sy = j;
      }

    leftDFS(sx, sy, 1);
    rightDFS(sx, sy, 1);
    BFS(sx, sy);
    cout << leftStep + 1 << ' ' << rightStep + 1 << ' ' << shortStep + 1
         << endl;
  }
  system("pause");
  return 0;
}