POJ-1258-Agri Ned

Posted 2021-02-07 ydddd

链接：https://vjudge.net/problem/POJ-1258#author=fuxianda

题意：

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. 

思路：

最小生成树

代码：

#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;
}