zoukankan      html  css  js  c++  java
  • HDU1162-Eddy's picture(最小生成树)

    Problem Description

    Eddy begins to like painting pictures recently ,he is sure of himself to become a painter.Every day Eddy draws pictures in his small room, and he usually puts out his newest pictures to let his friends appreciate. but the result it can be imagined, the friends are not interested in his picture.Eddy feels very puzzled,in order to change all friends 's view to his technical of painting pictures ,so Eddy creates a problem for the his friends of you.
    Problem descriptions as follows: Given you some coordinates pionts on a drawing paper, every point links with the ink with the straight line, causes all points finally to link in the same place. How many distants does your duty discover the shortest length which the ink draws?

    Input

    The first line contains 0 < n <= 100, the number of point. For each point, a line follows; each following line contains two real numbers indicating the (x,y) coordinates of the point.
    Input contains multiple test cases. Process to the end of file.

    Output

    Your program prints a single real number to two decimal places: the minimum total length of ink lines that can connect all the points.

    Sample Input

    3
    1.0 1.0
    2.0 2.0
    2.0 4.0

    Sample Output

    3.41

    Author

    eddy

    Recommend

    JGShining

    这是一个最小生成树的模板题目

    下面的代码用了Kruskal算法

    #include <cstdio>
    #include <vector>
    #include <algorithm>
    #include <cmath>
    
    using namespace std;
    
    const int N = 105;
    
    struct Edge
    {
        int x, y;
        double w;
    };
    
    struct Point
    {
        double x;
        double y;
    };
    
    int pre[N];
    Point point[N];
    Edge edges[N * N / 2];
    
    int i_p, i_e, cnt;
    double res;
    
    int root(int x)
    {
        if (x != pre[x])
        {
            pre[x] = root(pre[x]);
        }
        return pre[x];
    }
    
    bool merge(int x, int y)
    {
        int fx = root(x);
        int fy = root(y);
    
        bool ret = false;
    
        if (fx != fy)
        {
            pre[fx] = pre[fy];
            ret = true;
            --cnt;
        }
        return ret;
    }
    
    void init(int n)
    {
        cnt = n;
        res = 0;
    
        for (int i = 0; i <= n; ++i)
        {
            pre[i] = i;
        }
    }
    
    bool cmp(const Edge &a, const Edge &b)
    {
        return a.w < b.w;
    }
    
    int main()
    {
        int n;
        double dx, dy;
    
        while (scanf("%d", &n) != EOF)
        {
            init(n);
    
            i_e = i_p = 0;
    
            for (int i = 0; i < n; ++i)
            {
                scanf("%lf %lf", &dx, &dy);
                point[i_p].x = dx;
                point[i_p].y = dy;
                ++i_p;
            }
    
            for (int i = 0; i < n; ++i)
            {
                for (int j = i + 1; j < n; ++j)
                {
                    edges[i_e].x = i;
                    edges[i_e].y = j;
                    double dd = (point[i].x - point[j].x) * (point[i].x - point[j].x);
                    dd += (point[i].y - point[j].y) * (point[i].y - point[j].y);
                    edges[i_e].w = sqrt(dd);
                    ++i_e;
                }
            }
    
            sort(edges, edges + i_e, cmp);
    
            //the cnt == 1 indicates that the mixnum spanning tree is builded sucessfully.
            for (int i = 0; i < i_e && cnt != 1; ++i)
            {
                if (merge(edges[i].x, edges[i].y))res += edges[i].w;
            }
    
            printf("%.2lf
    ", res);
        }
        return 0;
    }
  • 相关阅读:
    张艾迪(创始人):视觉计算极简主义的设计
    张艾迪(创始人):同一个世界.同一个梦想
    张艾迪(创始人):Hello.世界...
    张艾迪(创始人):理念是全世界都在用....
    张艾迪(创始人):解码互联网天才
    张艾迪(创始人):艾迪成长记
    张艾迪(创始人): 趣味励志
    张艾迪(创始人): 励志的路上
    张艾迪(创始人):创始人故事无限N个
    张艾迪(创始人): 从诞生那一刻.走向整个世界
  • 原文地址:https://www.cnblogs.com/liuzhanshan/p/6234988.html
Copyright © 2011-2022 走看看