标签:i++ his net std this stream 并且 end nod
有n个农场,已知这n个农场都互相相通,有一定的距离,现在每个农场需要装光纤,问怎么安装光纤能将所有农场都连通起来,并且要使光纤距离最小,输出安装光纤的总距离
任意两个村庄之间的距离小于 100,000.
最小生成树
#include <iostream> #include <memory.h> #include <string> #include <istream> #include <sstream> #include <vector> #include <stack> #include <algorithm> #include <map> #include <queue> #include <math.h> #include <cstdio> using namespace std; typedef long long LL; const int MAXM = 10000+10; const int MAXN = 100+10; struct Node { double _x,_y; }node[MAXN]; struct Path { int _l,_r; double _value; bool operator < (const Path & that)const{ return this->_value < that._value; } }path[MAXM]; int Father[MAXN]; double a[MAXN]; int n, m, v; int s, p; int pos;//边数 int Get_F(int x) { return Father[x] = (Father[x] == x) ? x : Get_F(Father[x]); } void Init(int x) { for (int i = 1;i <= x;i++) Father[i] = i; } double Get_Len(Node a,Node b) { return sqrt((a._x - b._x) * (a._x - b._x) + (a._y - b._y) * (a._y - b._y)); } int main() { while (cin >> n) { Init(n); pos = 0; for (int i = 1;i <= n;i++) { for (int j = 1;j <= n;j++) { cin >> v; path[++pos]._l = i; path[pos]._r = j; path[pos]._value = v; } } sort(path + 1,path + 1 + pos); LL res = 0; for (int i = 1;i <= pos;i++) { int tl = Get_F(path[i]._l); int tr = Get_F(path[i]._r); if (tl != tr) { Father[tl] = tr; res += path[i]._value; } } cout << res << endl; } return 0; }
标签:i++ his net std this stream 并且 end nod
原文地址:https://www.cnblogs.com/YDDDD/p/10340316.html