POJ2560_TOJ2371Freckles

Freckles

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6195 Accepted: 3108

Description

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 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

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

31.0 1.02.0 2.02.0 4.0

Sample Output

3.41

Source

Waterloo local 2000.09.23

12045361wangwenhao2560Accepted296K0MSC++1671B2013-08-27 23:17:4912045351wangwenhao2560Wrong Answer  G++1717B2013-08-27 23:15:07

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <cmath>using namespace std;void kruscal(int edgeSize);const int maxn = 110;const int maxm = 51000;const double eps = 1e-8;int gn;struct point {    double dx;    double dy;}p[maxn];struct edge {    int x, y;    double w;}e[maxm];int f[maxn];double dist(double x, double y) {    return sqrt(x*x + y*y);//先不开方.}bool cmp(const edge& a, const edge& b) {    return a.w < b.w;}void build() {//建图.    int i, j;    int index = 0;    for(i = 1; i <= gn; i++) {        for(j = 1; j <= gn; j++) {           if(i == j) continue;           else {                e[index].x = i;                e[index].y = j;                e[index].w = dist(p[i].dx-p[j].dx, p[i].dy-p[j].dy);                index++;           }        }    }    kruscal(index);}int getfather(int x) {    if(x == f[x]) return x;    else return f[x] = getfather(f[x]);}void kruscal(int edgeSize) {    int i;    sort(e, e+edgeSize, cmp);    int cnt = gn;    double ans = 0.0;    for(i = 1; i <= gn; i++) f[i] = i;    for(i = 0; i < edgeSize; i++) {        int t1 = getfather(e[i].x);        int t2 = getfather(e[i].y);        if(t1 != t2) {            f[t1] = t2;            cnt--;            ans += e[i].w;        }        if(cnt == 1) break;    }    printf("%.2lf\n", ans);}int main(){    int i;    while(scanf("%d", &gn) != EOF) {        for(i = 1; i <= gn; i++) {            scanf("%lf%lf", &p[i].dx, &p[i].dy);        }        build();    }    return 0;}