ZOJ 1203 Swordfish （经典MST ~ Kruscal）Boruvka算法

链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=203

Description： 

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

分析：

1. 对图中各顶点，将与其关联，具有最小权边的边选入MST，得到由MST子树构成的森林；

2. 在图中陆续选择可以连接两颗不同子树且具有最小权值的边，将子树合并，最终构造成MST；

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <cmath>
#define MAXN 105  //顶点个数的最大值；
#define MAXM 5005  //边的个数的最大值；
#include <algorithm>
#define INF 0x1f1f1f1f
#define RST(N)memset(N, 0, sizeof(N))
using namespace std;

struct Edge
{
    int u, v;   //边的顶点；
    double w;   //边的权值；
}edge[MAXM];

int parent[MAXN], n, m, cas = 1;  //每个顶点对应的父亲节点， 顶点个数，边个数，测试数据个数；
double X[MAXN], Y[MAXN], res;     //每个顶点的X坐标和Y坐标；生成树的权值；

double Mul(double x) { return (double)x*x; }

void Init()
{
    for(int i=0; i<n; i++) scanf("%lf %lf", &X[i], &Y[i]);   //读入顶点个数；
    int k = 0;
    for(int i=0; i<n; i++) {
        for(int j=i+1; j<n; j++) {
            double dist = sqrt(Mul(X[i]-X[j]) + Mul(Y[i]-Y[j]));
            edge[k].u = i, edge[k].v = j, edge[k++].w = dist;
        }
    }
    m = k;
}

void Make_Set()    //初始化每个节点；
{
    for(int i=0; i<n; i++) {
        parent[i] = -1;
    }
}

int Find(int x)    //查找父亲节点；
{
    int s;
    for(s=x; parent[s]>=0; s=parent[s]);
    while(s != x) {
        int temp = parent[x];
        parent[x] = s;
        x = temp;
    }
    return s;
}

void Union_Set(int x, int y)   //节点合并；
{
    x = Find(x), y = Find(y);
    int temp = parent[x] + parent[y];
    if(parent[x] > parent[y]) {
        parent[x] = y;
        parent[y] = temp;
    }else {
        parent[y] = x;
        parent[x] = temp;
    }
}

int cmp(const void *a, const void *b)  //按边从小到大进行排序；
{
    Edge p1 = *(const Edge *)a;
    Edge p2 = *(const Edge *)b;
    if(p1.w > p2.w) return 1;
    else return -1;
}

void Kruscal()    //Kruscal 算法；
{ 
    int cnt = 0, Mc, Md;
    Make_Set();
    for(int i=0; i<m; i++) {
        Mc = edge[i].u, Md = edge[i].v;
        if(Find(Mc) != Find(Md)) {
            res += edge[i].w, cnt++;
            Union_Set(Mc, Md);
        }
        if(cnt >= n-1) break;
    }
}

int main()
{
    while(~scanf("%d", &n) && n) {
        Init();
        qsort(edge, m, sizeof(Edge), cmp);
        res = 0.0;    
        Kruscal();
        if(cas > 1) printf("\n");
        printf("Case #%d:\n", cas++);
        printf("The minimal distance is: %.2lf\n", res);
    }
    return 0;
}