HDU1162 Eddy's picture

题意大致是：

给出一组坐标点，每两个坐标点之间都可以连线，求最短的连线将所有点串在一起。

用普利姆算法AC····

模板题啊！····   -.-

#include<iostream>
#include<math.h>
using namespace std;

#define MAXN 110
#define INF 0x3fffff

struct Point {
	double x;
	double y;
}points[MAXN];
double map[MAXN][MAXN];
double dis[MAXN];	//存放现阶段到此点最断的里程
bool visited[MAXN];

double caldistance(Point x, Point y) {
	return (double)sqrt(pow((double)(x.x - y.x), 2) + pow((double)(x.y - y.y), 2));
}

double Prim(int n) {
	visited[0] = true;
	double min_dis = 0;
	dis[0] = 0;

	for (int i=1; i<n; i++) {
		visited[i] = false;
		dis[i] = map[0][i];
	}

	for (int i=1; i<n; i++) {	//N个点有N-1条边，故从1开始循环
		double temp_dis = INF;
		int dis_index = 0;
		for (int j=0; j<n; j++) {

			//寻找从j点出发那条路径最短;
			if (!visited[j] && temp_dis > dis[j]) {
				temp_dis = dis[j];
				dis_index = j;
			}		
		}
		visited[dis_index] = true;
		min_dis += temp_dis;

		for (int k=0; k<n; k++) {	//更新dis[];
			if (!visited[k] && map[k][dis_index] < dis[k]) {
				dis[k] = map[k][dis_index];
			}
		}
	}

	return min_dis;

}
int main() {
	int n;
	while (cin>>n) {
		for (int i=0; i<n; i++) {
			cin>>points[i].x;
			cin>>points[i].y;
		}
		for (int i=0; i<n; i++) {
			for (int j=0;j<n;j++) {
				map[i][j] = map[j][i] = caldistance(points[i], points[j]);
			}
		}

		printf("%.2lf\n",Prim(n));
	}

}