BFS+预处理 POJ - 3669 

向右看齐 2022-06-17 08:09 128阅读 0赞

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 (Xi, Yi) (0 ≤ Xi ≤ 300; 0 ≤ Yi ≤ 300) at time Ti (0 ≤ Ti ≤ 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: Xi, Yi, and Ti  
  
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

#include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <string>
    #include <map>
    #include <queue>
    #include <set>
    #include <algorithm>
    #define LL long long
    #define INF 0x7fffffff
    #define MAX 200010
    using namespace std;
    int n,x,y,time;
    int maps[305][305];
    int visit[305][305];
    int dir[5][2]={
       {-1,0},{1,0},{0,-1},{0,1},{0,0}};  //包含本身的位置
    struct point
    {
        int x;
        int y;
        int time;
    };
    int BFS()
    {
        if(maps[0][0]==0) return -1;  //maps[0][0]==0表示零时刻有炸弹爆炸 那么这个人就挂了
        if(maps[0][0]==-1) return 0;  //maps里面是-1代表这个点永远是安全的  不用移动了
        memset(visit,0,sizeof(visit));
        visit[0][0]=1;
        point p;
        p.x=0; p.y=0;p.time=0;
        queue<point> que;
        que.push(p);
        while(!que.empty())
        {
            point before=que.front();
            que.pop();
            if(maps[before.x][before.y]==-1) return before.time;
            else{
                for(int k=0;k<4;k++)
                {
                    point next;
                    next.x=before.x+dir[k][0];
                    next.y=before.y+dir[k][1];
                    next.time=before.time+1;
                    if( next.x>=0 &&next.y>=0 && !visit[next.x][next.y])
                    {
                       if(maps[next.x][next.y]==-1) return next.time;
                       else
                       {
                           if(maps[next.x][next.y]>next.time)
                           {
                               visit[next.x][next.y]=1;
                               que.push(next);
                           }
                       }
                    }
                }
            }
        }
        return -1;
    }
    int main()
    {
       scanf("%d",&n);
       memset(maps,-1,sizeof(maps));
       for(int i=1;i<=n;i++)
       {
           scanf("%d%d%d",&x,&y,&time);
             for(int k=0;k<5;k++)
             {
                 int xx=x+dir[k][0];
                 int yy=y+dir[k][1];
                 if(xx>=0 && yy>=0)
                 {
                     if(maps[xx][yy]==-1)
                        maps[xx][yy]=time;
                     else
                        maps[xx][yy]=min(maps[xx][yy],time);
                 }
             }
    
       }
        printf("%d\n",BFS());
        return 0;
    }
    
    
    /*解题思路(参考大神的）：
    
    首先对矩阵进行预处理，先将矩阵置为-1表示从没有流星雨来临，
    然后将流星雨来临的点及其周围四格置为流星雨降临的时间，
    同一格有两次以上的流星雨时取时间最早的，因为求最短路不会走走过的点。
    然后从(0，0)开始BFS。原点处的值为-1则说明已经是安全的地方不需要移动，值为0说明开始就挂了。
    用maps[i][j] > next.t 就可以判断这个点是否可以在当前时间暂时访问那*/
    
    
    
    /*题意：
    
     巨大流星雨即将袭来。每个流星会对击中的地方以及周围（上下左右四格）造成破坏。
    Bessie开始时位于（0, 0）位置，并希望逃到一处不会被袭击到的地方（在第一象限内）。
    已知每移动一格需要1个时间单位，被流星破坏后的地方不能再进入。给出M个流星在T时击中的地方（X, Y），
    问Bessie能否逃到安全的地方，若能输出最短时间，否则输出-1  */