(step6.1.3)hdu 1875(畅通工程再续——最小生成树)-优快云博客

本文链接：https://blog.youkuaiyun.com/hjd_love_zzt/article/details/10338411

本文探讨了如何运用最小生成树算法解决特定问题，包括将二维点转化为一维点的过程，以及如何筛选有效的边进行计算。通过具体代码实现，展示了算法在实际问题中的应用，最终通过kruscal算法求解最小生成树。

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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");
		}
	}
}