习题7-18(uva-10384)

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这是一个使用C++编写的矩阵迷宫寻路算法，通过深度优先搜索（DFS）解决从起点到终点的最短路径问题。代码中涉及了标准库的多种数据结构如栈、队列、地图表示以及位操作来检查路径可行性。

#include <iostream>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <cstring>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <numeric>
#include <chrono>
#include <ctime>
#include <cmath>
#include <cctype>
#include <string>
#include <cstdio>
#include <iomanip>


#include <thread>
#include <mutex>
#include <condition_variable>
#include <functional>
#include <iterator>
using namespace std;
const int Row = 6, Col = 8, D2P[4] = { 'W','N','E','S' }, Dirs[4][2] = { {0,-1},{-1,0 },{0,1},{1,0} };
enum Dir {West = 0,North = 1,East = 2,South = 3};

int matrix[Row][Col] = {0}, sx, sy,vis[Row][Col];
string ans;

bool CheckDir(int input, int d) {
	return input & (1 << d);
}
int h(int x, int y)
{
	int res = 10000;
	for (int i = 1; i <= 4; ++i)
	{
		if (!CheckDir(matrix[i][1], West))
			res = min(res, abs(x - i) + y);
		if (!CheckDir(matrix[i][6], East))
			res = min(res, abs(x - i) + 7 - y);
	}
	for (int i = 1; i <= 6; ++i)
	{
		if (!CheckDir(matrix[1][i], North))
			res = min(res, abs(y - i) + x);
		if (!CheckDir(matrix[4][i], South))
			res = min(res, abs(y - i) + 5 - x);
	}
	return res;
}

bool dfs(int x, int y, int depth,string& temp) {
	if (depth == 0 && (x == 0 || y == 0 || x == 5 || y == 7)) {
		ans = temp;
		return true;
	}
	else if (depth <= 0) return false;

	if (h(x, y)  > depth) return false;

	for (int i = 0;i<4;i++)
	{
		int nx = x + Dirs[i][0];
		int ny = y + Dirs[i][1];
		if (vis[nx][ny]) continue;

		if (!CheckDir(matrix[x][y], i)) {
			vis[nx][ny] = 1;
			temp.push_back(D2P[i]);
			if (dfs(nx, ny, depth - 1, temp)) return true;
			temp.pop_back();
			vis[nx][ny] = 0;
		}else if (!CheckDir(matrix[nx][ny], i) && (nx!= 0 && nx != 5 && ny != 0 && ny != 7))
		{
			vis[nx][ny] = 1;
			temp.push_back(D2P[i]);
			matrix[nx][ny] ^= 1 << i;
			matrix[x][y] ^= 1 << i;
			matrix[nx + Dirs[i][0]][ny + Dirs[i][1]] ^= 1 << ((i + 2)%4);
			if (dfs(nx, ny, depth - 1, temp)) return true;
			matrix[nx][ny] ^= 1 << i;
			matrix[x][y] ^= 1 << i;
			matrix[nx + Dirs[i][0]][ny + Dirs[i][1]] ^= 1 << ((i + 2) % 4);
			temp.pop_back();
			vis[nx][ny] = 0;
		}
	}
	return false;
}
int main()
{
	while (cin >> sy >> sx && sx && sy) {
		for (int i = 1; i <= 4; i++) {
			for (int j = 1; j <= 6; j++)
				cin >> matrix[i][j];
		}
		string temp = "";
		memset(vis, 0, sizeof(vis));
		vis[sx][sy] = 1;
		for (int i  = 1;; i++)
		{
			if (dfs(sx, sy, i, temp)) break;
		}
		
		cout << ans << endl;
	}
}