ZCMU-训练赛-Problem A: Freckles

Problem A: Freckles

Time Limit: 1 Sec   Memory Limit: 128 MB
Submit: 41   Solved: 21
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Description

Problem A: Freckles

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.

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.

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

Input

Output

Sample Input

3

1.0 1.0

2.0 2.0

2.0 4.0

Sample Output

3.41

【解析】

这道题的意思其实就是叫我们求能连通这几个点的，最短的直线，其实就是最小生成树的问题了，只是最小生成树是给你

点和权重，这里其实给的是距离，我们把距离算出来之后，再进行筛选就比较好做了这里用的是prim算法。

#include
#include
#include
#include
#include
using namespace std;
#define INF 0x7fffffff
double graph[101][101],distances[101];
int visit[101];
int n;
double prim(int x)
{
    int i,j,mid;
    double sum=0,min1;
    for(i=0;idistances[j]&&visit[j]==0)//筛选出未到过的点当中哪个距离最小
            {
                min1=distances[j];
                mid=j;
            }
        }
        visit[mid]=1;
        sum+=min1;//加上最小距离
        for(j=0;jgraph[mid][j]&&visit[j]==0)//更新从mid这个点到其余的点
            {
                distances[j]=graph[mid][j];
            }
        }
    }
    return sum;
}
int main()
{
    int i,j,k;
    while(~scanf("%d",&n))
    {
        double a[101],b[101];
        memset(graph,0,sizeof(graph));
        for(i=0;i