hdu1242 Rescue--BFS

原题链接： http://acm.hdu.edu.cn/showproblem.php?pid=1242

一：题意

x代表卫兵，a代表终点，r代表起始点，.代表路，#代表墙
路花费一秒，x花费两秒
问到达终点的最少时间
思路：BFS+优先队列

二：AC代码

#define _CRT_SECURE_NO_DEPRECATE 
#define _CRT_SECURE_Cy1_OVERLOAD_STANDARD_NAMES 1 

#include
#include 
#include 
#include 
#include 
using namespace std;

struct Node
{
	int x, y;
	int step;
	bool operator<(const Node & node) const
	{
		return node.step < step;
	}
};

char ch[205][205];
int dis[205][205];
int dir[4][2] = { 1,0,-1,0,0,1,0,-1 };
int sx, sy;//起点，也就是朋友的位置
int dx, dy;
int n, m;

bool bfs()
{
	Node t, tt;
	priority_queue pq;
	t.x = sx;
	t.y = sy;
	t.step = 0;
	pq.push(t);
	dis[sx][sy] = 0;

	while (!pq.empty())
	{
		t = pq.top();
		pq.pop();
		if (t.x == dx&&t.y == dy)
		{
			printf("%d\n", t.step);
			return true;
		}

		for (int i = 0; i < 4; i++)
		{
			tt = t;
			tt.x += dir[i][0];
			tt.y += dir[i][1];
			if (tt.x < 0 || tt.y < 0 || tt.x >= n || tt.y >= m || ch[tt.x][tt.y] == '#')
				continue;

			tt.step++;
			if (ch[tt.x][tt.y] == 'x')
				tt.step++;

			if (dis[tt.x][tt.y] >= tt.step)
			{
				dis[tt.x][tt.y] = tt.step;
				pq.push(tt);
			}
		}
	}

	return false;
}

int main() 
{
	while (~scanf("%d%d", &n, &m))
	{
		for (int i = 0; i < n; i++)
		{
			getchar();
			for (int j = 0; j < m; j++)
			{
				dis[i][j] = 99999999;
				scanf("%c", &ch[i][j]);
				if (ch[i][j] == 'a')
				{
					dx = i;
					dy = j;
				}
				else if (ch[i][j] == 'r')
				{
					sx = i;
					sy = j;
				}
			}
		}
		
		if (!bfs())
			printf("Poor ANGEL has to stay in the prison all his life.\n");
	}
	return 0;
}