POJ 3669 Meteor Shower

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本文介绍了一道经典的路径寻找问题——POJ3669流星雨问题，并提供了完整的广度优先搜索(BFS)算法解决方案。通过预处理每个格点被破坏的时间，并使用队列进行节点的遍历，最终找到Bessie能够到达的安全位置及所需最短时间。

POJ 3669 Meteor Shower

题目链接

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

Sample Output

分析

广度优先搜索BFS。BFS模板利用队列完成。main()函数中对数据做了预处理。

源代码

#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
int visit[1000][1000];
int dx[5]={-1,0,0,0,1};
int dy[5]={0,1,0,-1,0};
struct point{
    int x,y;
    int step;
};
queue<point> que;
int bfs(){
    if(visit[0][0]==0) return -1;
    if(visit[0][0]==-1) return 0;
    point s,cur,next;
    s.x=0;//起点
    s.y=0;
    s.step=0;
    que.push(s);
    while(que.size()){
        cur=que.front();
        que.pop();
        for(int i=0;i<5;i++){
            next.x=cur.x+dx[i];
            next.y=cur.y+dy[i];
            next.step=cur.step+1;

            if(next.x>=0 && next.y>=0 && next.x<1000 && next.y<1000){
                if(visit[next.x][next.y]==-1){
                    return next.step;
                }
                if(next.step<visit[next.x][next.y]){
                    visit[next.x][next.y]=next.step;
                    que.push(next);
                }
            }

        }
    }
    return -1;
}

int main(){
    int M,x,y,t,xx,yy;
    cin>>M;
    memset(visit,-1,sizeof(visit));
    for(int i=0;i<M;i++){
        cin>>x>>y>>t;
        for(int j=0;j<5;j++){
            xx=x+dx[j];
            yy=y+dy[j];
            if(xx>=0 && yy>=0 && xx<1000 && yy<1000){
                if(visit[xx][yy]==-1){
                    visit[xx][yy]=t;
                }
                else{
                    visit[xx][yy]=min(visit[xx][yy],t);
                }
            }
        }
    }
    int result=bfs();
    cout<<result<<endl;
    return 0;  
}