POJ-1287-Networking

链接：https://vjudge.net/problem/POJ-1287#author=dream2017

题意：

存在许多点和点与点之间的路径，路径长度不一，点到点之间可能存在多条路径。挑选部分路径使得所有点连通且总路径长度最小。

思路:

Kruskal

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>
using namespace std;
typedef long long LL;
const int MAXM = 10000;
const int MAXN = 100;
struct Path
{
    int _l;
    int _r;
    int _value;
    bool operator < (const Path & that)const {
        return this->_value < that._value;
    }
}path[MAXM];

int Father[MAXN];

int Get_F(int x)
{
    return Father[x] = (Father[x] == x) ? x : Get_F(Father[x]);
}

int main()
{
    int p,r;
    while (~scanf("%d%d",&p,&r) && p)
    {
        for (int i = 1;i<=p;i++)
            Father[i] = i;
        for (int i = 1;i<=r;i++)
            scanf("%d%d%d",&path[i]._l,&path[i]._r,&path[i]._value);
        sort(path+1,path+1+r);
        int sum = 0;
        for (int i = 1;i<=r;i++)
        {
            int tl = Get_F(path[i]._l);
            int tr = Get_F(path[i]._r);
            if (tl != tr)
            {
                Father[tl] = tr;
                sum += path[i]._value;
            }
        }
        printf("%d
",sum);
    }

    return 0;
}