HDU - 1102 Prim &（ kruskal 并查集 优先队列)

HDU - 1102 Prim &（ kruskal 并查集 优先队列)

Prim

只需要解决已经联通的道路这一个问题（这里可以将已联通的城市之间的距离设置为0）

#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
	int n, m, map[105][105], Min[105];
	bool B[105];
	while (cin >> n)
	{
		int MST = 0;
		memset(B, true, sizeof(B));
		memset(Min, 0x3f3f3f3f, sizeof(Min));
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				scanf("%d", &map[i][j]);
		cin >> m;
		int t1, t2;
		for (int i = 0; i < m; i++)
		{
			scanf("%d%d", &t1, &t2);
			map[t1][t2] = map[t2][t1] = 0;//解决已联通的问题
		}
		int k;
        /*******prim主体********/
		Min[1] = 0;
		for (int i = 1; i <= n; i++)
		{
			k = 0;
			for (int j = 1; j <= n; j++)
			{
				if (B[j] && Min[j] < Min[k])
					k = j;
			}
			MST += Min[k];
			B[k] = false;
			for (int j = 1; j <= n; j++)
			{
				if (B[j] && Min[j] > map[j][k])
					Min[j] = map[j][k];
			}
		}
        /*******************/
		cout << MST << endl;
	}
	return 0;

kruskal使用优先队列存储可以将n^2的复杂度缩小到nlogn！！此外还可以用vector和邻接矩阵

#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
struct Node 
{
	int u, v, w;//u,v是边的两个顶点，w为边长（权值）
};

int pre[110];//用于并查集

struct tmp//重写
{
	bool  operator() (Node a, Node b)
	{
		return a.w > b.w; 
	}
};

int find(int x)//路径压缩优化的find
{
	if (x != pre[x])
		pre[x] = find(pre[x]);
	return pre[x];
}

void merge(int a, int b)//合并结点
{
	int t1 = find(a);
	int t2 = find(b);
	if (t1 != t2)
		pre[t1] = t2;
}

int main()
{
	int n, m;
	while (cin >> n)
	{
		priority_queue <Node, vector<Node>, tmp > q;//定义一个优先队列
		int ans = 0;
		for (int i = 0; i <= n; i++)
			pre[i] = i;
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
			{
				Node t;
				t.u = i; t.v = j;
				cin >> t.w;
				q.push(t);
			}
		cin >> m;
		int t1, t2;
		for (int i = 0; i < m; i++)
		{
			cin >> t1 >> t2;
			if (find(t1) != find(t2))
				merge(t1, t2);
		}
		while (!q.empty())
		{
			int t1 = find(q.top().u);
			int t2 = find(q.top().v);
			if (t1!=t2)
			{
				ans += q.top().w;
				merge(t1,t2);
			}
			q.pop();
		}
		cout << ans << endl;
	}
	return 0;
}