A place to record sth.

题意：给出三维坐标系上的一些球的球心坐标和其半径，搭建通路，使得他们能够相互连通。如果两个球有重叠的部分则算为已连通，无需再搭桥。求搭建通路的最小费用（费用就是边权，就是两个球面之间的距离）。思路：把球A和球B看做无向图图的两个结点，则边权 = AB球面距离 = A球心到B球心的距离 –  A球半径 – B球半径，注意若边权#include#include#include#inc

思路：最小生成树模板。#include#include#include#define min(a,b) a<b?a:busing namespace std;const int INF=0x3f3f3f3f;const int MAXN=110;bool vis[MAXN];int lowc[MAXN];//点是 0 n-1int Prim(int cost[][MAX

思路：最小生成树模板。#include#include#includeusing namespace std;const int INF=0x3f3f3f3f;const int MAXN=110;bool vis[MAXN];int lowc[MAXN];//点是 0 n-1int Prim(int cost[][MAXN],int n){ int ans=0;

思路：最小生成树模板题。#include#include#include#includeusing namespace std;const int INF=0x3f3f3f3f;const int MAXN=110;bool vis[MAXN];double lowc[MAXN];//点是 0 n-1double Prim(double cost[][MAXN],int

思路：最小生成树模板题。#include#include#includeusing namespace std;const int INF=0x3f3f3f3f;const int MAXN=110;bool vis[MAXN];int lowc[MAXN];//点是 0 n-1int Prim(int cost[][MAXN],int n){ int ans=0

Happy ValueTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1286    Accepted Submission(s): 380Problem DescriptionIn an apartment