Meteor Shower

最新推荐文章于 2024-02-25 14:21:32 发布

winer00

最新推荐文章于 2024-02-25 14:21:32 发布

阅读量200

点赞数

CC 4.0 BY-SA版权

分类专栏：算法与数据结构

本文链接：https://blog.youkuaiyun.com/qq_41485193/article/details/79179859

算法与数据结构专栏收录该内容

42 篇文章

订阅专栏

面对即将到来的流星雨威胁，Bessie需要找到一条安全的路径到达一个永远不受流星撞击的地方。通过在坐标平面上避开被破坏的位置，Bessie必须确定能够到达安全地点所需的最短时间。

Bessie hears that an extraordinary meteor shower is coming; reports say that these meteors will crash into earth and destroy anything they hit. Anxious for her safety, she vows to find her way to a safe location (one that is never destroyed by a meteor) . She is currently grazing at the origin in the coordinate plane and wants to move to a new, safer location while avoiding being destroyed by meteors along her way.

The reports say that M meteors (1 ≤ M ≤ 50,000) will strike, with meteor i will striking point (X_i, Y_i) (0 ≤ X_i≤ 300; 0 ≤ Y_i≤ 300) at time T_i (0 ≤ T_i ≤ 1,000). Each meteor destroys the point that it strikes and also the four rectilinearly adjacent lattice points.

Bessie leaves the origin at time 0 and can travel in the first quadrant and parallel to the axes at the rate of one distance unit per second to any of the (often 4) adjacent rectilinear points that are not yet destroyed by a meteor. She cannot be located on a point at any time greater than or equal to the time it is destroyed).

Determine the minimum time it takes Bessie to get to a safe place.

Input

* Line 1: A single integer: M
* Lines 2..M+1: Line i+1 contains three space-separated integers: X_i, Y_i, and T_i

Output

* Line 1: The minimum time it takes Bessie to get to a safe place or -1 if it is impossible.

Sample Input

Sample Output

#include<stdio.h>  
#include<string.h>  
#include<math.h>  
#include<queue>  
#include<algorithm>  
#define inf 0x3f3f3f3f
using namespace std;  
int vis[350][350];
int vis1[330][330];
const int dis[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
typedef struct node{
	int x,y,t;
}Q;
void BFS()  
{   
    Q  n1,temp;
    queue<node>Q;
    n1.x=0;n1.y=0;n1.t=0;
    Q.push(n1);
    int x1,y1,i;
	if(vis[0][0]==0) {
	    printf("-1\n");
	return ; 
    }
	if(vis[0][0]==inf) {
	    printf("0\n");
	return ; 
     }
     vis1[0][0]=1;
    while(!Q.empty()){
    	n1=Q.front();
    	Q.pop();
    	if(vis[n1.x][n1.y]==inf)
    	 {
    	 	 printf("%d\n",n1.t);
    	 	 return ;
		 }
    	for(i=0;i<4;i++)
    	{
		    temp=n1;
    		x1=temp.x+dis[i][0];
    		y1=temp.y+dis[i][1];
			temp.t=temp.t+1;
			temp.x=x1;
    		temp.y=y1;
    		if(!vis1[x1][y1]&&x1>=0&&x1<=301&&y1>=0&&y1<=301&&temp.t<vis[x1][y1])
    		{
    			
    			if(vis[n1.x][n1.y]==inf)
					 {
					 	 printf("%d\n",n1.t);
					 	 return ;
					 }
    			vis1[x1][y1]=1;
    		
    			Q.push(temp);
			}
		}
	}
	printf("-1\n");
}  
int main()  
{  
    int m;
    while(~scanf("%d",&m))
    {	
	    int i,j;
    	int a,b,c;
    	int a1,b1;
    	memset(vis1,0,sizeof(vis1));
    	for(i=0;i<=310;i++)
    	    for(j=0;j<=310;j++)
    	        vis[i][j]=inf;
    	for(i=0;i<m;i++){
            scanf("%d%d%d",&a,&b,&c);
            if(vis[a][b]>c)
                vis[a][b]=c;
            for(j=0;j<4;j++)
            {
             	a1=a+dis[j][0];
             	b1=b+dis[j][1];
                if(a1>=0&&b1>=0&&vis[a1][b1]>c)
                     vis[a1][b1]=c;
			 }     
		}
    	BFS();
	}
    return 0;
}