HOJ 1301 Jungle Roads (最小生成树)

本题只是输入看起来比较复杂，细心一点处理好就可以了。
用三个一维数组代替二维数组存储边，降低时间复杂度。

#include<iostream>
#include<cstring>
using namespace std;
const int N = 30, NN = 80, INF = 99999999;
int c1[NN], c2[NN], c3[NN], f[N];//三个一维数组代替二位数组存储数据，降低时间复杂度

int find(int x)
{//寻找根节点
	if (x == f[x]) return x;
	f[x] = find(f[x]);
	return f[x];
}

int mer(int x, int y)
{//将x和y所在的两个集合合并
	x = find(x);
	y = find(y);
	if (x != y)
		f[x] = y;
	return 1;
}

int main()
{
	int n, x, y, k, num, sum;
	char xx, yy;
	while (cin >> n && n)
	{
		num = 0;
		sum = 0;
		memset(c1, 0, sizeof(c1));
		memset(c2, 0, sizeof(c2));
		memset(c3, 0, sizeof(c3));
		for (int i = 0; i <= N; i++)
			f[i] = i;//顶点初始化
		for (int i = 0; i < n - 1; i++)
		{
			cin >> xx >> k;
			x = xx - 'A';
			while (k--)
			{
				cin >> yy;
				y = yy - 'A';
				c1[num] = x;
				c2[num] = y;
				cin >> c3[num];
				num++;
			}
		}//for
		while (1)
		{
			int min0 = INF, flag = 1;
			for(int i = 0; i < num; i++)
				if (c3[i] < min0 && find(c1[i]) != find(c2[i]))
				{
					min0 = c3[i];
					x = i;
					flag = 0;
				}
			if (flag) break;
			sum += min0;
			mer(c1[x], c2[x]);
		}
		cout << sum << endl;
	}
	return 0;
}

来源：CSDN

作者：qq_44724908

链接：https://blog.csdn.net/qq_44724908/article/details/104188644