(step6.1.3)hdu 1875(畅通工程再续——最小生成树)

题目大意:本题是中文题，可以直接在OJ上看

解题思路:最小生成树

1)本题的关键在于把二维的点转化成一维的点

 

for (i = 0; i < n; ++i) {

			scanf("%d%d", &point[i].x, &point[i].y);

			point[i].id = i;

		}

2)可用边的计算

 

 

int count = 0;

		for (i = 0; i < n; ++i) {

			for (j = i + 1; j < n; ++j) {

				double distances = getDistance(point[i],point[j]);



				if (distances >= 10.0 && distances <= 1000.0) {

					e[count].begin = point[i].id;

					e[count].end = point[j].id;

					e[count].weight = distances;

					count++;

				}

			}

		}

 

代码如下:

 

/*

 * 1875_3.cpp

 *

 *  Created on: 2013年8月26日

 *      Author: Administrator

 */



#include <iostream>

#include <math.h>

using namespace std;



struct edge {

	int begin;

	int end;

	double weight;

};



struct Point {

	int x;

	int y;

	int id;

};



const int maxn = 6000;

int father[maxn];

edge e[maxn];



int find(int a) {

	if (a == father[a]) {

		return a;

	}



	father[a] = find(father[a]);

	return father[a];

}



double kruscal(int count) {

	int i;

	double sum = 0;

	for (i = 0; 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;

}



double getDistance(const Point& a, const Point& b) {

	return sqrt(

			(double) ((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y)));

}



int main() {

	int t;

	scanf("%d", &t);

	while (t--) {

		int n;

		scanf("%d", &n);

		int i, j;

		memset(father, 0, sizeof(father));

		for (i = 0; i < n; ++i) {

			e[i].begin = -1;

			e[i].end = -1;

			e[i].weight = 0;

		}

		Point point[n + 1];



		for (i = 0; i < n; ++i) {

			scanf("%d%d", &point[i].x, &point[i].y);

			point[i].id = i;

		}



		int count = 0;

		for (i = 0; i < n; ++i) {

			for (j = i + 1; j < n; ++j) {

				double distances = getDistance(point[i],point[j]);



				if (distances >= 10.0 && distances <= 1000.0) {

					e[count].begin = point[i].id;

					e[count].end = point[j].id;

					e[count].weight = distances;

					count++;

				}

			}

		}



		sort(e,e + count,compare);

		double result = kruscal(count);



		int num = 0;

		for (i = 0; i < n; ++i) {

			if (father[i] == i) {

				num++;

			}

		}



		if (num == 1) {

			printf("%.1lf\n", result * 100);

		} else {

			printf("oh!\n");

		}

	}

}