(step 6.1.9)hdu 1162(Eddy's picture——最小生成树)

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题目大意：输入一个整数n表示表示有n个点。在接下来的n行中，每行有两个整数x ， y 。分别表示一个点的横坐标以及纵坐标。求距离最小的连线

解题思路:
1)二维----->>一维

for(i = 1 ; i <= n ; ++i){scanf("%lf%lf",&point[i].x,&point[i].y);point[i].id = i;}

2)求所有连线的长度

int count = 0;for(i = 1 ; i <= n ;  ++i){for(j = i+1 ; j <= n ; ++j){e[count].begin = point[i].id;e[count].end = point[j].id;e[count].weight = getDistance(point[i],point[j]);count++;}}

因为之前写过一篇博客已对这种题有较详细的解释，所以这里就不详细解释了

代码如下：

/* * 1162_1.cpp * *  Created on: 2013年8月27日 *      Author: Administrator */#include <iostream>#include <cmath>#include <algorithm>using namespace std;struct edge{int begin;int end;double weight;};const int maxn = 600;int father[maxn];edge e[maxn*maxn];int find(int x){if( x == father[x]){return x;}father[x] = find(father[x]);return father[x];}double kruscal(int count){int i;double sum = 0;for(i = 1 ; i < maxn ; ++i){father[i] = i;}for(i = 0 ; i < count ; ++i){int fx = find(e[i].begin);int fy = find(e[i].end);if(fx != fy){father[fx] = fy;sum += e[i].weight;}}return sum;}bool compare(const edge& a , const edge& b){return a.weight < b.weight;}struct Point{double x;double y;int id;};Point point[maxn];double getDistance(const Point& p1 , const Point& p2){return sqrt((p1.x - p2.x)*(p1.x - p2.x) + (p1.y - p2.y)*(p1.y - p2.y) );}int main(){int n;while(scanf("%d",&n)!=EOF){int i,j;for(i = 1 ; i <= n ; ++i){scanf("%lf%lf",&point[i].x,&point[i].y);point[i].id = i;}int count = 0;for(i = 1 ; i <= n ;  ++i){for(j = i+1 ; j <= n ; ++j){e[count].begin = point[i].id;e[count].end = point[j].id;e[count].weight = getDistance(point[i],point[j]);count++;}}sort(e , e + count , compare);double sum = kruscal(count);printf("%.2lf\n",sum);}}