Wheels(bfs遍历,CERC 2014)

原题链接:http://acm.hnu.cn/online/?action=problem&type=show&id=13397

对于每一个已经开始运动的轮子要求它周围和他相接触的轮子的运动状态,然后让那些轮子依次进入队列= =

速度关系== 高中学过物理应该都知道吧

  1 #include<stdio.h>
  2 #include<string.h>
  3 #include<math.h>
  4 #include<algorithm>
  5 #include<iostream>
  6 #include<queue>
  7 using namespace std;
  8 #define maxn 2000
  9 struct node
 10 {
 11     int x,y,r,rate1,rate2;
 12     int flag;
 13 };
 14 int hash[maxn];
 15 node a[maxn];
 16 queue<node>q;
 17 int  dis(node a,node b)
 18 {
 19     int f=0;
 20     double len=pow(a.x-b.x,2.0)+pow(a.y-b.y,2.0);
 21     if(len-pow(a.r+b.r,2.0)<=0.00000001)
 22     f=1;
 23     return f;
 24 }
 25 int gcd(int x,int y)
 26 {
 27     int ans;
 28     if(y!=0)
 29     ans=gcd(y,x%y);
 30     else
 31     ans=x;
 32     return ans;
 33 }
 34 int main()
 35 {
 36     int t,n;
 37     scanf("%d",&t);
 38     while(t--)
 39     {
 40         while(!q.empty())
 41          q.pop();
 42         memset(hash,0,sizeof(hash));
 43         scanf("%d",&n);
 44         for(int i=0;i<n;i++)
 45         {
 46             scanf("%d %d %d",&a[i].x,&a[i].y,&a[i].r);
 47             a[i].flag=0;
 48         }
 49         a[0].flag=1;
 50         a[0].rate1=1;
 51         a[0].rate2=1;
 52         q.push(a[0]);
 53         while(!q.empty())
 54         {
 55             node now=q.front();
 56             q.pop();
 57             for(int i=1;i<n;i++)
 58             {
 59                 if(hash[i])
 60                 continue;    
 61                 node b;
 62                 if(dis(a[i],now)&&hash[i]==0)
 63                 {
 64                     if(now.flag==1)
 65                     a[i].flag=-1;
 66                     else if(now.flag==-1)
 67                     a[i].flag=1;
 68                         b.x=a[i].x;
 69                         b.y=a[i].y;
 70                         b.r=a[i].r;
 71                         hash[i]=1;
 72                         a[i].rate1=now.rate1*now.r;
 73                         a[i].rate2=now.rate2*a[i].r;
 74                          int cnt=gcd(a[i].rate1,a[i].rate2);
 75                           a[i].rate1/=cnt;
 76                           a[i].rate2/=cnt;
 77                         q.push(a[i]);
 78                     }
 79                 }
 80             }
 81         for(int i=0;i<n;i++)
 82         {
 83             if(a[i].flag==0)
 84             {
 85                 printf("not moving
");
 86             }
 87             else if(a[i].flag==1)
 88             {
 89                 if(a[i].rate2==1)
 90                 printf("%d clockwise
",a[i].rate1);
 91                 else
 92                 printf("%d/%d clockwise
",a[i].rate1,a[i].rate2);
 93             }
 94             else
 95             {
 96                 if(a[i].rate2==1)
 97                 printf("%d counterclockwise
",a[i].rate1);
 98                 else
 99                 printf("%d/%d counterclockwise
",a[i].rate1,a[i].rate2);
100             }
101         }
102         }
103     return 0;
104 }
原文地址:https://www.cnblogs.com/NaCl/p/4735598.html