POJ 3669 Meteor Shower BFS求最小时间

Meteor Shower

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 31358		Accepted: 8064

Description

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

4
0 0 2
2 1 2
1 1 2
0 3 5

Sample Output

5

 

INPUT DETAILS:
There are four meteors, which strike points (0, 0); (2, 1); (1, 1); and (0, 3) at times 2, 2, 2, and 5, respectively.


                                                                               
    t = 0                t = 2              t = 5
5|. . . . . . .     5|. . . . . . .     5|. . . . . . .    
4|. . . . . . .     4|. . . . . . .     4|# . . . . . .   * = meteor impact
3|. . . . . . .     3|. . . . . . .     3|* # . . . . .  
2|. . . . . . .     2|. # # . . . .     2|# # # . . . .   # = destroyed pasture
1|. . . . . . .     1|# * * # . . .     1|# # # # . . .   
0|B . . . . . .     0|* # # . . . .     0|# # # . . . .   
  --------------      --------------      -------------- 
  0 1 2 3 4 5 6       0 1 2 3 4 5 6       0 1 2 3 4 5 6 
 

Sample Output 

5

OUTPUT DETAILS:
Examining the plot above at t=5, the closest safe point is (3, 0) -- but Bessie's path to that point is too quickly blocked off by the second meteor. The next closest point is (4,0) -- also blocked too soon. Next closest after that are lattice points on the
(0,5)-(5,0) diagonal. Of those, any one of (0,5), (1,4), and (2,3) is reachable in 5 timeunits.



       5|. . . . . . .   
       4|. . . . . . .   
       3|3 4 5 . . . .    Bessie's locations over time
       2|2 . . . . . .    for one solution
       1|1 . . . . . .   
       0|0 . . . . . .   
         -------------- 
         0 1 2 3 4 5 6  

题意：Bessie从原点出发，而后有N个流星会在某个时刻落下，它们会破坏砸到的这个方格还会破坏

四边相邻的方块，输出多少时间以后他能够到达安全的地方。若是可能，输出最优解，不可能则输出-1。

 

#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
#include<queue>
using namespace std;
int dir[5][2]={{0,0},{0,-1},{0,1},{1,0},{-1,0}};//原点停留（处理爆炸点），上下左右
int a[505][505];
int n,m,cnt;
struct node
{
    int x;
    int y;
    int t;
}temp,now;

int check(int x,int y)
{
    if(x>=0&&x<500&&y>=0&&y<500)
        return 1;
    else
        return 0;
}

int bfs()
{
    if(a[0][0]==0)//刚开始走就被炸死了
        return -1;
    if(a[0][0]==-1)//起点就是安全的地方，不用走了
        return 0;

    temp.x=0,temp.y=0,temp.t=0;//起点
    queue<node>p;
    p.push(temp);

    while(!p.empty())
    {
        now=p.front();
        p.pop();
        for(int i=1;i<5;i++)//不能原地停留
        {
            int dx,dy,dt;
            dx=now.x+dir[i][0];
            dy=now.y+dir[i][1];
            dt=now.t+1;
            if(check(dx,dy)==0)//走出边界
                continue;
            if(a[dx][dy]==-1)//到达安全区域
                return dt;
            if(dt>=a[dx][dy])//走进爆炸区域
                continue;
            a[dx][dy]=dt;//更新时间
            temp.x=dx;
            temp.y=dy;
            temp.t=dt;
            p.push(temp);//下一个搜索的点进队列
        }
    }
    return -1;//到不了安全区域
}
//bfs是从最近的点开始搜索，因此获得的第一个答案就是最近
int main()
{
    cin>>n;
    memset(a,-1,sizeof(a));//初始化地图全部地方均可以走
    while(n--)
    {
        int x,y,t;
        cin>>x>>y>>t;
        for(int i=0;i<5;i++)//预处理全部爆炸结束以后的地图
        {
            int dx,dy;
            dx=x+dir[i][0];
            dy=y+dir[i][1];
            if(check(dx,dy)==0)//超出地图范围
                continue;
            if(a[dx][dy]==-1)
                a[dx][dy]=t;
            else//if(a[dx][dy]!=-1)
                a[dx][dy]=min(a[dx][dy],t);//取最早爆炸的时间
        }
    }
    cout<<bfs()<<endl;

    return 0;
}