ZOJ 1203 Swordfish（Prim算法求解MST）

题目：

There exists a world within our world
A world beneath what we call cyberspace.
A world protected by firewalls,
passwords and the most advanced
security systems.
In this world we hide
our deepest secrets,
our most incriminating information,
and of course, a shole lot of money.
This is the world of Swordfish.

We
all remember that in the movie Swordfish, Gabriel broke into the World
Bank Investors Group in West Los Angeles, to rob $9.5 billion. And he
needed Stanley, the best hacker in
the world, to help him break into the password protecting the bank
system. Stanley's lovely daughter Holly was seized by Gabriel, so he had
to work for him. But at the last moment, Stanley made some little trick
in his hacker mission: he injected a trojan
horse in the bank system, so the money would jump from one account to
another account every 60 seconds, and would continue jumping in the next
10 years. Only Stanley knew when and where to get the money. If Gabriel
killed Stanley, he would never get a single
dollar. Stanley wanted Gabriel to release all these hostages and he
would help him to find the money back.
  You
who has watched the movie know that Gabriel at last got the money by
threatening to hang Ginger to death. Why not Gabriel go get the money
himself? Because these money keep jumping,
and these accounts are scattered in different cities. In order to
gather up these money Gabriel would need to build money transfering
tunnels to connect all these cities. Surely it will be really expensive
to construct such a transfering tunnel, so Gabriel
wants to find out the minimal total length of the tunnel required to
connect all these cites. Now he asks you to write a computer program to
find out the minimal length. Since Gabriel will get caught at the end of
it anyway, so you can go ahead and write the
program without feeling guilty about helping a criminal.

Input:
The
input contains several test cases. Each test case begins with a line
contains only one integer N (0 <= N <=100), which indicates the
number of cities you have to connect. The next
N lines each contains two real numbers X and Y(-10000 <= X,Y <=
10000), which are the citie's Cartesian coordinates (to make the problem
simple, we can assume that we live in a flat world). The input is
terminated by a case with N=0 and you must not print
any output for this case.

Output:
You
need to help Gabriel calculate the minimal length of tunnel needed to
connect all these cites. You can saftly assume that such a tunnel can be
built directly from one city to another.
For each of the input cases, the output shall consist of two lines: the
first line contains "Case #n:", where n is the case number (starting
from 1); and the next line contains "The minimal distance is: d", where d
is the minimal distance, rounded to 2 decimal
places. Output a blank line between two test cases.

Sample Input:

5
0 0
0 1
1 1
1 0
0.5 0.5
0

Sample Output:

Case #1:
The minimal distance is: 2.83

题意描述：
题目描述的很有意思（大部分都是跟题无关的废话），简单来说给你N个点的坐标，让你计算它们的最小生成树的距离。
解题思路：
将数据转化成邻接矩阵，使用Prim算法即可。
代码实现：

 #include<stdio.h>
 #include<math.h>
 #include<string.h>
 struct n
 {
     double x,y;
     int find;
 };
 int main()
 {
     int n,i,j,book[],count,k,t=;
     double e[][],dis[],sum,min;
     struct n c[];
     while(scanf("%d",&n),n != )
     {
         for(i=;i<=n;i++)
             scanf("%lf%lf",&c[i].x,&c[i].y);
         for(i=;i<=n;i++)
         {
             for(j=i;j<=n;j++)
             {
                 if(i==j)
                 e[i][j]=;
                 else
                 {
                     e[i][j]=sqrt((c[i].x-c[j].x)*(c[i].x-c[j].x)+(c[i].y-c[j].y)*(c[i].y-c[j].y));
                     e[j][i]=e[i][j];
                 }
             }
         }
         memset(book,,sizeof(book));
         for(i=;i<=n;i++)
             dis[i]=e[][i];
         book[]=;
         sum=;//sum 初始化
         count=;//count 初始化
         count++;
         while(count < n)
         {
             min=;
             for(i=;i<=n;i++)
             {
                 if(!book[i] && dis[i]<min)
                 {
                     min=dis[i];
                     j=i;
                 }
             }
             book[j]=;
             count++;
             sum += dis[j];
             for(k=;k<=n;k++)
             {
                 if(!book[k] && dis[k] > e[j][k])
                     dis[k]=e[j][k];
             }
         }

         if(t != )
         printf("\n");
         printf("Case #%d:\nThe minimal distance is: %.2lf\n",++t,sum);

     }
     return ;
 } 

易错分析：

1、很无奈，初始化问题要牢记。

2、格式问题