uva-705-深搜

题意,就是根据斜线组成的迷宫,判断能够组成多少个闭环.

解法:

放大俩倍或者三倍

俩倍

    ------->10

　　　　　　01

三倍

　　------->100

　　　　　　 010

　　　　　　 001

然后深搜,有个问题,没有判断是否成环,它竟然过了,待证明----它到底对不对

#include<stdio.h>
#include<iostream>
#include<queue>
#include<memory.h>
using namespace std;
const int MAX = 80 * 3;
int maps[MAX][MAX];
int r, c;
void dfs(int x, int y, int tr, int tc, int* total, int* ok)
{
	*total = *total + 1;
	maps[x][y] = 1;
	if(x == 0 || x == (tr - 1) || y == 0 || y == (tc - 1))
		*ok = 0;
	if(x > 0 && maps[x - 1][y] == 0)
		dfs(x - 1, y, tr, tc, total, ok);
	if(y > 0 && maps[x][y - 1] == 0)
		dfs(x, y - 1, tr, tc, total, ok);
	if(x < (tr - 1) && maps[x + 1][y] == 0)
		dfs(x + 1, y, tr, tc, total, ok);
	if(y < (tc - 1) && maps[x][y + 1] == 0)
		dfs(x, y + 1, tr, tc, total, ok);

}

int main()
{
	freopen("d:\1.txt", "r", stdin);
	int T = 1;
	while (cin >> c >> r)
	{
		if(r == 0 && c == 0)
			return 0;
		memset(maps, 0, sizeof(maps));
		char cc;
		for(int i = 0; i < r; i++)
		{
			for(int j = 0; j < c; j++)
			{
				cin >> cc;
				if(cc == '/')
				{
					maps[3 * i][3 * j + 2] = 1;
					maps[3 * i + 1][3 * j + 1] = 1;
					maps[3 * i + 2][3 * j] = 1;
				}
				else if(cc == '\')
				{
					maps[3 * i][3 * j] = 1;
					maps[3 * i + 1][3 * j + 1] = 1;
					maps[3 * i + 2][3 * j + 2] = 1;
				}
			}
		}
		int mr = 3 * r;
		int mc = 3 * c;
		int t = 0;
		int max = 0;
		for(int i = 0; i < mr; i++)
			for(int j = 0; j < mc; j++)
			{
				if(maps[i][j] == 0)
				{
					int ok = 1;
					int total = 0;
					dfs(i, j, mr, mc, &total, &ok);
					if(ok)
					{
						max = max > total ? max : total;
						t++;
					}
				}
			}

		cout << "Maze #" << T << ":" << endl;
		if(t == 0)
			cout << "There are no cycles." << endl;
		else
			cout << t << " Cycles; the longest has length " << max / 3 << "."
					<< endl;
		cout << endl;
		T++;

	}
	return 0;
}