[bfs] poj 3669 Meteor Shower

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 (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

先预处理会爆炸的区域，设vis[i][j] 的值为这块区域上最早爆炸的时间，不会爆炸的区域为-1，然后做bfs搜索，将可以走到的路径推入队列，直到第一次遇见永远安全的区域为止(vis[][] = -1)

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <vector>#include <queue>using namespace std;int vis[1005][1005];int dx[5] = {1,-1,0,0,0};int dy[5] = {0,0,1,-1,0};int n,m;struct node{    int x,y,step;};queue<node> Q;int bfs(int x,int y){    if (vis[0][0] == -1)  return 0;    if (vis[0][0] == 0) return -1;    node a,next;    a.x = x;    a.y = y;    a.step = 0;    Q.push(a);    while(!Q.empty())    {        a =Q.front();        Q.pop();        for(int i = 0; i < 5;i++)        {            int nx = a.x + dx[i];            int ny = a.y + dy[i];            next.step = a.step + 1;            if( nx>=0 && nx<=1000 && ny>=0 && ny<=1000 )            {                if (vis[nx][ny]==-1)                {                    return next.step;                }                if (next.step < vis[nx][ny])                {                    next.x = nx,next.y = ny;                    vis[nx][ny] = next.step;                    Q.push(next);                }            }        }    }    return -1;}int main(){    while(cin>>m)    {        int x,y,t,nx,ny;        memset(vis,-1,sizeof vis);        for(int i = 0; i <m;i++)        {            cin>>x>>y>>t;            for(int i = 0;i<5;i++)            {                nx = x + dx[i];                ny = y + dy[i];                if (nx>=0 && nx<=1000 && ny>=0 && ny<=1000)                {                    if (vis[nx][ny] == -1)                        vis[nx][ny] = t;                    else                        vis[nx][ny] = min(vis[nx][ny],t);                    //先处理完会爆炸的区域                }            }        }        printf("%d\n",bfs(0,0));    }    return 0;}