POJ

Meteor Shower

 POJ - 3669 

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

40 0 22 1 21 1 20 3 5

Sample Output

5

题意：

    要下流星雨，给个整数M，表示有M颗流星，接下来有M组数据，每组数据三个正整数，前两个表示这颗流星落下的坐标（坐标全在第一象限内），第三个表示落下的时间，流星落下时会将自身坐标和周围上下左右四个点总共五个位置给炸毁。Bessie在时间为零时从坐标原点出发，每单位时间内可以往她周围上下左右四个方向走一步（当然也是在第一象限内），求她到达安全地点（就是不会被流星炸毁的地方）的最短时间。如果不能够找到一个安全的地方，输出“-1”。

解题思路：

    求最短时间，应该想到用广度搜索，这题其实就是在基础广搜的基础上，对地图进行一个标记，标记流星炸毁的最早的时间，更新流星炸毁时间时，图内更新的应该是流星将这一地区炸毁的最早时间。广搜时判断一下到达这个点时是否流星已经将这个地点炸毁，若没有炸毁，将这个点加入队列继续搜索，直到搜索到一个从始至终没有被流星炸毁的地方，就表明Bessie到达了安全的地方。

注意：

    刚开始应该将地图初始化一下，初始化流星炸毁的时间，但是不能初始化为0，因为时间是从0开始，可能有些流星在时间为0时就炸毁了。若判断坐标是否出界，上限不是300，起码得是303，因为可能坐标包含300的流星会将坐标为301的地方炸毁，Bessie需要走到坐标为302的地方才算安全。流星炸毁的地点包括它本身的位置。还有就是可能坐标原点就是安全的地点，在进入广搜时应该单独判断一下。

AC代码：

#include<stdio.h>#include<string.h>#include<queue>#include<algorithm>using namespace std;int mp[500][500], minn, flag;int book[500][500];int nx[5][2]= {{0,-1},{-1,0},{0,1},{1,0},{0,0}};struct node       //存入流星炸毁的坐标，时间{    int x;    int y;    int temp;};void bang( int x, int y, int t ) //标记流星炸毁的坐标和更新最早的炸毁时间{    for( int i = 0; i < 5; i++ ) //记得还有本身也需要炸毁    {        int tx = x + nx[i][0];        int ty = y + nx[i][1];        if( tx<0 || ty<0 )       //若判断上限应该起码为303            continue;        if( mp[tx][ty] == -1 )            mp[tx][ty] = t;        else            mp[tx][ty] = min(t,mp[tx][ty]);//保存流星炸毁的最早时间    }}void bfs( ){    queue<node>Q;    node now, stp;    now.x = 0;    //初始坐标为0 0    now.y = 0;    now.temp = 0; //时间也是从0开始    Q.push(now);    while( !Q.empty() )    {        now = Q.front();        Q.pop();        if( mp[now.x][now.y] == -1 )        {            flag = 1;            minn = now.temp;   //若原点就是安全的地点            return ;           // 记录步数直接返回        }        for( int i = 0; i < 4; i++ )        {            int tx = now.x + nx[i][0];            int ty = now.y + nx[i][1];            if( tx<0 || ty<0 )                continue;            if( mp[tx][ty] == -1 )  //找到安全的地点            {                flag = 1;                minn = now.temp + 1;                return ;            }            if( now.temp+1 < mp[tx][ty] && !book[tx][ty] )            {                        //可以走的坐标                book[tx][ty] = 1;                stp.x = tx;                stp.y = ty;                stp.temp = now.temp + 1;                Q.push(stp);            }        }    }}int main(){    int m;    while( ~scanf("%d",&m) )    {        int x1, y1, t1;        memset(mp,-1,sizeof(mp));        for( int i = 0; i < m; i++ )        {            scanf("%d%d%d",&x1,&y1,&t1);            bang(x1,y1,t1);   //标记流星爆炸        }        flag = 0;        memset(book,0,sizeof(book));        book[0][0] = 1;        bfs();        if( flag )            printf("%d\n",minn);        else            printf("-1\n");    }    return 0;}