原題鏈接 Meteor Shower Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 18098 Accepted: 4704 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: MLines 2..M+1: Line i+1 contains three space-separated integers: Xi, Yi, and TiOutput
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 Source
USACO 2008 February Silver
#include <cstdio>#include <cstring>#include <iostream>#include <queue>using namespace std;const int INF = 0x3f3f3f3f;const int dx[4]={0,0,-1,1};const int dy[4]={-1,1,0,0};const int maxn=400 + 10;int a[maxn][maxn],d[maxn][maxn];//隕石最早到達的時間和到達該點最短的時間int n;typedef struct node{ int x,y;}node;int bfs(){ if(a[0][0]==0) return -1; queue<node> que; que.push((node){0,0}); d[0][0]=0; while(!que.empty()){ node now=que.front(); que.pop(); int nowx=now.x,nowy=now.y; if(a[nowx][nowy]==INF) return d[nowx][nowy]; for(int i=0;i<4;i++){ int nx=nowx+dx[i],ny=nowy+dy[i]; if(nx>=0 && ny>=0 && d[nx][ny]==INF && d[nowx][nowy] + 1 < a[nx][ny]){ d[nx][ny]=d[nowx][nowy]+1; que.push((node){nx,ny}); } } } return -1;}int main(){ while(scanf("%d",&n)==1){ for(int i=0;i<maxn;i++){ fill(a[i],a[i]+maxn,INF); fill(d[i],d[i]+maxn,INF); } int x,y,t; while(n--){ scanf("%d%d%d",&x,&y,&t); a[x][y]=min(a[x][y],t); for(int i=0;i<4;i++){ int nx=x+dx[i],ny=y+dy[i]; if(nx>=0 && ny>=0) a[nx][ny]=min(a[nx][ny],t); } } int res=bfs(); cout << res << endl; }}新聞熱點
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