1913 Problem B Freckles

问题 B: Freckles
时间限制: 1 Sec 内存限制: 32 MB
献花: 26 解决: 0
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题目描述
In an episode of the Dick Van Dyke show, little Richie connects the freckles on his Dad’s back to form a picture of the Liberty Bell. Alas, one of the freckles turns out to be a scar, so his Ripley’s engagement falls through.
Consider Dick’s back to be a plane with freckles at various (x,y) locations. Your job is to tell Richie how to connect the dots so as to minimize the amount of ink used. Richie connects the dots by drawing straight lines between pairs, possibly lifting the pen between lines. When Richie is done there must be a sequence of connected lines from any freckle to any other freckle.
输入
The first line contains 0 < n <= 100, the number of freckles on Dick’s back. For each freckle, a line follows; each following line contains two real numbers indicating the (x,y) coordinates of the freckle.
输出
Your program prints a single real number to two decimal places: the minimum total length of ink lines that can connect all the freckles.
样例输入
3
2723.62 7940.81
8242.67 11395.00
4935.54 6761.32
9
10519.52 11593.66
12102.35 2453.15
7235.61 10010.83
211.13 4283.23
5135.06 1287.85
2337.48 2075.61
6279.72 13928.13
65.79 1677.86
5324.26 125.56
0
样例输出
8199.56
32713.73

此代码在codeup上无法通过，可能是精度的问题，但是在牛客网上可顺利通过，在牛客网上跑需要将主函数中的while(cin >> N,N)替换为 while(cin >> N)

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <cfloat>
#include <iomanip>
#include <map>

#define INF INT32_MAX
using namespace std;

struct Edge
{
    int u, v;
    double cost;
};

struct vex
{
    double x;
    double y;
    vex(double _x, double _y) :x(_x), y(_y) {};
};

const int MaxN = 110;
const int MaxL = 5000;
int Father[MaxN];
Edge edge[MaxL];
vector<vex> Vex;

int N,L;

bool cmp(Edge a, Edge b)
{
    return a.cost < b.cost;
}

int findRoot(int s)
{
    while (s != Father[s])
        s = Father[s];

    return s;
}

double kruskal(int s)
{
    sort(edge, edge + L,cmp);
    for (int i = 0; i < N; ++i)
        Father[i] = i;

    int curNum = 0;
    double Mincost = 0;
    for (int i = 0; i < L; ++i)
    {
        int u = edge[i].u;
        int v = edge[i].v;
        int ur = findRoot(u);
        int vr = findRoot(v);
        if (ur != vr)
        {
            Father[vr] = ur;
            Mincost += edge[i].cost;;
            ++curNum;

            if (curNum == N - 1)break;
        }
    }

    if (curNum != N - 1)return -1;
    return Mincost;

}


int main()
{
#ifdef _DEBUG
    freopen("data.txt", "r+", stdin);
#endif // _DEBUG

    std::ios::sync_with_stdio(false);

    while (cin >> N, N)
    {
        double x, y;
        Vex.clear(); L = 0;

        for (int i = 0; i < N; ++i)
        {
            cin >> x >> y;
            for (int k = 0; k < Vex.size(); ++k)
            {
                edge[L].u = i;
                edge[L].v = k;
                edge[L++].cost = sqrt(pow(Vex[k].x - x, 2) + pow(Vex[k].y - y, 2));
            }
            Vex.push_back(vex(x, y));
        }

        cout << setiosflags(ios::fixed) << setprecision(2) << kruskal(0) << endl;
    }

    return 0;
}

牛客上通过截图：