bzoj1615 / P2903 [USACO08MAR]麻烦的干草打包机The Loathesome Hay Baler

P2903 [USACO08MAR]麻烦的干草打包机The Loathesome Hay Baler

细节题。$O(n^{2})$的$bfs$可过。

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<cctype>
 #include<cmath>
 #define re register
 using namespace std;
 typedef double db;
 void read(int &x){
     char c=getchar();x=;int f=;
     while(!isdigit(c)) f&=(c!='-'),c=getchar();
     while(isdigit(c)) x=(x<<)+(x<<)+(c^),c=getchar();
     x=f?x:-x;
 }
 #define N 1052
 const db eps=1e-;
 int sqr(int a){return a*a;}
 struct node{int x,y,r;}a[N];
 int n,xt,yt,st,ed;
 db d[N],v[N]; bool vis[N];
 db dist(node A,node B){return sqrt((db)(sqr(A.x-B.x)+sqr(A.y-B.y)));}
 void dfs(int x){
     if(x==ed){
         cout<<(int)d[x];
         exit();
     }//终点直接跳出
     db tmp=v[x]*(db)a[x].r;
     for(int i=;i<=n;++i){
         if(i==x||vis[i]) continue;
         if(fabs(dist(a[x],a[i])-(db)(a[x].r+a[i].r))<eps){//距离实时计算：两个齿轮相切
             vis[i]=; v[i]=tmp/(db)a[i].r;
             d[i]=d[x]+v[i]; dfs(i);
         }
     }
 }
 int main(){
     read(n);read(xt);read(yt);
     for(re int i=;i<=n;++i){
         read(a[i].x);read(a[i].y);read(a[i].r);
         if(a[i].x==xt&&a[i].y==yt) ed=i;
         if(a[i].x==&&a[i].y==) st=i;
     }v[st]=d[st]=;vis[st]=;dfs(st);
 }