clas creat const lag sam sin ppr 報告 ech

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? 譯文：埃迪最近開始喜歡畫畫，他肯定自己要成為一名畫家。埃迪每天都在他的小房間裏畫畫，他通常會拍出他最新的照片讓他的朋友們欣賞。但結果可想而知，朋友對他的照片不感興趣。埃迪感到非常困惑，為了將所有朋友的觀點轉變為他的繪畫技術，埃迪為他的朋友們創造了一個問題。問題描述如下：給出你在繪圖紙上的一些坐標信息，每一點用直線與油墨連接，使所有點最終連接在同一個地方。你有多少距離發現了墨水吸取的最短長度？ 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. 譯文：第一行包含0 <n <= 100，即點數。對於每一點，一條線跟隨; 每個以下行包含兩個實數，指示該點的（x，y）坐標。輸入包含多個測試用例。處理到文件的結尾。 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. 譯文：你的程序打印一個單一的實數到小數點後兩位：可以連接所有點的墨水線的最小總長度。 Sample Input 3 1.0 1.0 2.0 2.0 2.0 4.0 Sample Output 3.41 解題思路：最小生成樹，簡單題。 AC代碼之Prim算法：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 105;
 4 int n;
 5 bool vis[maxn];
 6 double lowdist[maxn],dist[maxn][maxn];
 7 pair<double,double> point[maxn];
 8 double vecx(double x1,double y1,double x2,double y2){
 9     return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
10 }
11 double Prim(){
12     for(int i=1;i<=n;++i)
13         lowdist[i]=dist[1][i];
14     lowdist[1]=0;vis[1]=true;
15     double res=0.0;
16     for(int i=1;i<n;++i){
17         int k=-1;
18         for(int j=1;j<=n;++j)
19             if(!vis[j] && (k==-1||lowdist[k]>lowdist[j]))k=j;
20         if(k==-1)break;
21         vis[k]=true;
22         res+=lowdist[k];
23         for(int j=1;j<=n;++j)
24             if(!vis[j])lowdist[j]=min(lowdist[j],dist[k][j]);
25     }
26     return res;
27 }
28 int main()
29 {
30     while(cin>>n){
31         for(int i=1;i<=n;++i)
32             cin>>point[i].first>>point[i].second;
33         for(int i=1;i<=n;++i)
34             for(int j=1;j<=n;++j)
35                 dist[i][j]=vecx(point[j].first,point[j].second,point[i].first,point[i].second);
36         memset(vis,false,sizeof(vis));
37         cout<<setiosflags(ios::fixed)<<setprecision(2)<<Prim()<<endl;
38     }
39     return 0;
40 }

題解報告：hdu 1162 Eddy's picture