POJ Meteor Shower(BFS)
来源:互联网 发布:c语言数组作为形参 编辑:程序博客网 时间:2024/06/06 02:54
点击打开题目链接
Meteor ShowerTime Limit: 1000MSMemory Limit: 65536KTotal Submissions: 19992
Accepted: 5189
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
40 0 22 1 21 1 20 3 5
Sample Output
5
Source
USACO 2008 February Silver[Submit] [Go Back] [Status] [Discuss]
Home Page Go Back To top
题目大意:
有一场流星雨要降临,给定流星雨落得位置和时间,最初Bessie在坐标原点上,求她到达安全地点所需要的时间。流星雨落下来后除去落到当前坐标,还会覆盖相邻上下左右四个点。
思路:
每分钟更新地图的方法显然不可取,所以最初令地图上所有点为-1,之后改为时间值,大于这个时间值才可行。构图完后再用宽搜BFS做即可。
#include<iostream>#include<cstdio>#include<cstring>#include<queue>using namespace std;const int maxn=300+5;int map[maxn][maxn];int dx[]={0, 0, 1, 0, -1},dy[]={0, 1, 0, -1, 0};int x,y,t;int M;struct node{ int x, y, t;};int bfs(){ if(map[0][0]==-1)return 0; else if(map[0][0]==0) return -1; queue<node>Q; node temp,now; temp.x=temp.y=temp.t=0; Q.push(temp); while(!Q.empty()){ now=Q.front(); Q.pop(); for(int i=0;i<5;i++){ temp.x=now.x+dx[i]; temp.y=now.y+dy[i]; temp.t=now.t+1; if(temp.x>=0&&temp.x<maxn&&temp.y>=0&&temp.y<maxn){ if(temp.t<map[temp.x][temp.y]){ map[temp.x][temp.y]=temp.t; Q.push(temp); } else if(map[temp.x][temp.y]==-1){ return temp.t; } } } } return -1;}int main(){ while(~scanf("%d",&M)){ memset(map,-1,sizeof(map)); for(int i=0;i<M;i++){ scanf("%d%d%d",&x,&y,&t); for(int i=0;i<5;i++){ int tempx=x+dx[i]; int tempy=y+dy[i]; if(tempx>=0&&tempx<maxn&&tempy>=0&&tempy<maxn){ if(map[tempx][tempy]>t||map[tempx][tempy]==-1) map[tempx][tempy]=t; } } } printf("%d\n",bfs()); } return 0;}
- POJ Meteor Shower(BFS)
- POJ 3669 Meteor Shower (BFS)
- POJ 3669 Meteor Shower (BFS + 预处理)
- POJ 3669 Meteor Shower (bfs)
- POJ 3669 Meteor Shower(bfs)
- poj 3669 Meteor Shower(BFS)
- POJ 3669 Meteor Shower(BFS)
- POJ 3669-Meteor Shower(BFS)
- POJ - 3669 - Meteor Shower(bfs)
- POJ 3669 Meteor Shower(BFS)
- POJ 3669 Meteor Shower(BFS)
- POJ 3669 Meteor Shower(BFS)
- POJ 3669 Meteor Shower(bfs)
- poj 3669-Meteor Shower(简单bfs)
- POJ 3669 Meteor Shower(BFS)
- Meteor Shower (bfs)
- Meteor Shower (bfs)
- poj 3669 Meteor Shower (bfs)
- 一位10年Java工作经验的架构师聊Java和工作经验
- 【Maven实战】之依赖,仓库,镜像
- Mysql语句查询优化
- 思科认证入门级课程介绍(一)
- 数据库分片
- POJ Meteor Shower(BFS)
- Jenkins之——构建Java Maven项目(Jar)并发布到远程服务器
- Python面试题(原创)
- Top-Down Design
- ubuntu 终端安装 bpython
- PAT 1038. 统计同成绩学生(20)
- Git学习实用指南:什么是Git?
- image-loader 图片缓存
- 【Java】java 中的泛型通配符