【UVA 10034 Freckles】& Kruskal & 最小生成树

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.
Input
The input begins with a single positive integer on a line by itself indicating the number of the cases
following, each of them as described below. This line is followed by a blank line, and there is also a
blank line between two consecutive inputs.
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.
Output
For each test case, the output must follow the description below. The outputs of two consecutive cases
will be separated by a blank line.
Your program prints a single real number to two decimal places: the minimum total length of ink
lines that can connect all the freckles.
Sample Input
1
3
1.0 1.0
2.0 2.0
2.0 4.0
Sample Output
3.41

题意 ： 给出 n 个点的坐标，求把所有点都连接在一起的最小代价

思路 ： Kruskal

AC代码：

#include<cstdio>
#include<cmath>
#include<deque>
#include<queue>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAX = 1e5 + 10;
const int INF = 1e9 + 7;
typedef long long LL;
double x[110],y[110];
struct node{
    int a,b;
    double vl;
}st[110 * 110];
int f[110];
int fn(int x){ return f[x] == x ? f[x] : f[x] = fn(f[x]); }
void bc(int x,int y){
    int fx = fn(x),fy = fn(y);
    f[fy] = fx;
}
bool cmp(node i,node j) { return i.vl < j.vl; };
int main()
{
    int T,n;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        int nl = 0;
        for(int i = 1; i <= n; i++){
            f[i] = i;
            scanf("%lf %lf",&x[i],&y[i]);
            for(int j = 1; j < i; j++){
                double xx = x[i] - x[j];
                double yy = y[i] - y[j];
                node o;
                o.vl = sqrt(xx * xx + yy * yy);
                o.a = i,o.b = j;
                st[++nl] = o;
            }
        }
        sort(st + 1,st + nl + 1,cmp);
        double ans = 0;
        for(int i = 1; i <= nl; i++){
            int a = st[i].a,b =st[i].b;
            if(fn(a) != fn(b)) ans += st[i].vl,bc(a,b);
        }
        printf("%.2f\n",ans);
        if(T) puts("");
    }
    return 0;
}