[POJ] 3669 Meteor Shower [BFS]

最新推荐文章于 2021-09-18 12:12:18 发布

ctsas

最新推荐文章于 2021-09-18 12:12:18 发布

阅读量262

点赞数

CC 4.0 BY-SA版权

分类专栏： bfs 文章标签： poj bfs

本文链接：https://blog.youkuaiyun.com/ctsas/article/details/60144731

bfs 专栏收录该内容

13 篇文章

订阅专栏

面对即将到来的流星雨威胁，Bessie需要找到一条安全的路径到达一个不会被流星摧毁的地方。本篇介绍了如何使用广度优先搜索(BFS)算法来解决这一问题，并给出了具体的实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

题解

WA到差点哭出来
看了讨论区才知道，我没有考虑是会炸到相同格子

#include<stdio.h>
#include<string.h>
#include<queue>
#define MAX_N 303
using namespace std;
struct node{int x,y,s;};
int _map[MAX_N][MAX_N];
int M;
int ox[]={0,0,0,1,-1};
int oy[]={0,1,-1,0,0};

node make_node(int x,int y,int s){
    node t;
    t.x=x,t.y=y,t.s=s;
    return t;
}
int bfs(){
    queue<node> que;
    que.push(make_node(0,0,0));

    while(!que.empty()){
        node t=que.front();que.pop();
        for(int i=1;i<5;i++){
            int x=t.x+ox[i];
            int y=t.y+oy[i];
            if(x<0||y<0||x>=MAX_N||y>=MAX_N) continue;
            if(_map[x][y]==-1) return t.s+1;
            else if(t.s+1<_map[x][y]){
                que.push(make_node(x,y,t.s+1));
                _map[x][y]=t.s+1;
            }
        }
    }
    return -1;
}

int main()
{
    scanf("%d",&M);
        memset(_map,-1,sizeof(_map));
        for(int i=0,X,Y,t;i<M;i++){
            scanf("%d%d%d",&X,&Y,&t);
            for(int j=0;j<5;j++){
                int x=X+ox[j];
                int y=Y+oy[j];
                if(x<0||y<0||_map[x][y]!=-1&&t>=_map[x][y]) continue;
                _map[x][y]=t;
            }
        }
        int ans=bfs();
        printf("%d\n",ans);

    return 0;
}