# Eddy‘s picture

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7267    Accepted Submission(s): 3676

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
```

kruskal算法实现

```#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\n", res);
}
return 0;
}```

HDU 1162 Eddy's picture（图论-最小生成树）

## HDU 1162 Eddy&#39;s picture【最小生成树，Prime算法+Kruskal算法】

Eddy's picture Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9334    Accepted Submission(s): 4711 Problem Description Eddy begins to like painting pictures recently ,he is sure of himself to

## hdu 1162 Eddy&#39;s picture 最小生成树入门题 Prim+Kruskal两种算法AC

Eddy's picture Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7428    Accepted Submission(s): 3770 Problem Description Eddy begins to like painting pictures recently ,he is sure of himself to

## HDU 1162 Eddy&#39;s picture (最小生成树)(java版)

Eddy's picture 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1162 --每天在线,欢迎留言谈论. 题目大意: 给你N个点,求把这N个点连在一起的最短总距离. 思路: 假设每两两点之间都有路径,求最小生成树. AC代码:(Java) 1 import java.util.Scanner; 2 import java.math.*; 3 public class Main { 4 public static final int MAX

## 题解报告：hdu 1162 Eddy&#39;s picture

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 i