ZOJ 1203 Swordfish 剑鱼行动 最小生成树，Kruskal算法

题目链接：ZOJ 1203 Swordfish 剑鱼行动

Swordfish

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：给定平面上N个城市的位置，计算连接这N个城市所需线路长度总和的最小值。

分析：最小生成树，Kruskal问题求解。注意两个城市之间都有一条边相连。还有每两组输出之间空一行。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
using namespace std;

#define maxn 5055
double ans;
int cnt;
int parent[110];
struct edge
{
    int u, v;
    double w;
}EG[maxn];

bool cmp(edge a, edge b)
{
    return a.w < b.w;
}
int cmp2(const void *a, const void *b)
{
    edge aa = *(const edge*)a;
    edge bb = *(const edge*)b;
    if(aa.w > bb.w) return 1;
    return -1;
}
int Find(int x)
{
    if(parent[x] == -1) return x;
    return Find(parent[x]);
}
void Kruskal()
{
    memset(parent, -1, sizeof(parent));
    sort(EG, EG+cnt, cmp);
    //qsort(EG, cnt, sizeof(EG[0]), cmp2);
    ans = 0;
    for(int i = 0; i < cnt; i++)
    {
        int t1 = Find(EG[i].u), t2 = Find(EG[i].v);
        if(t1 != t2)
        {
            ans += EG[i].w;
            parent[t1] = t2;
        }
    }
}
int main()
{
    int n, cas = 0;
    double x[110], y[110];
    while(scanf("%d", &n), n)
    {
        for(int i = 0; i < n; i++)
            scanf("%lf%lf", &x[i], &y[i]);
        cnt = 0;
        for(int i = 0; i < n; i++)
            for(int j = i+1; j < n; j++)
            {
                EG[cnt].u = i;
                EG[cnt].v = j;
                EG[cnt].w = sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
                cnt++;
            }
        Kruskal();
        if(cas > 0)
            printf("\n");
        printf("Case #%d:\nThe minimal distance is: %.2f\n", ++cas, ans);
        //++cas;
    }
    return 0;
}