poj_3669_Meteor Shower(BFS+预处理)

http://poj.org/problem?id=3669

/*思路：BFS+预处理
先预处理会爆炸的区域，BFS，遇到-1则结束*/
#include <iostream>
#include <queue>
#include <algorithm>
using namespace std;
int visit[1001][1001];
int dx[5] = {0,0,1,0,-1};
int dy[5] = {0,1,0,-1,0};
typedef struct
{
	int x,y;
	int step;
}point;
queue<point> que;
int bfs(int x,int y)
{
	int cnt;
	if(visit[0][0] == 0)  return -1;//无法逃离
	if(visit[0][0] == -1) return 0;  //安全，无需逃离
	point s,cur,next;
	s.x = x;
	s.y = y;
	s.step = 0;
	que.push(s);
	while(!que.empty())
	{
		cur = que.front();
		que.pop();
		for(int i = 0;i < 5;i++)
		{
			next.x = cur.x + dx[i];
			next.y = cur.y + dy[i];
			next.step = cur.step + 1;

			if(next.x >= 0 && next.y >= 0 && next.y < 1001 && next.x < 1001)
			{
				if(visit[next.x][next.y] == -1)
				{
					return next.step;
				}
				if(next.step < visit[next.x][next.y])
				{
					visit[next.x][next.y] = next.step;//记录最小的步数
					que.push(next);
				}
			}
		}
	}
	return -1;
}
int main()
{

	int m;
    while(cin >> m)
	{
        int x,y,t,xx,yy;
	    memset(visit,-1,sizeof(visit));
	    for(int i = 0;i < m;i++)
	    {
	        cin >> x >> y >> t;

	        for(int i = 0;i <= 4;i++)
	        {
	            xx = x + dx[i];
	            yy = y + dy[i];
	            if(xx >= 0 && yy >= 0 && xx < 1001 && yy < 1001)//保证在有限区域内
				{
		            if(visit[xx][yy] == -1)
		            	 visit[xx][yy] = t;
		            else
		            	visit[xx][yy] = min(visit[xx][yy],t);
				}
	        }
	    }
	    cout << bfs(0,0) << "\n";
	}
    return 0;
}