【LittleXi】Minimum Spanning Tree

【LittleXi】Minimum Spanning Tree

Introduction

In graph theory algorithms, it is often necessary to find the smallest edge that can connect all nodes

kruskal algorithm

Algorithm Introduction

The algorithm uses the greedy idea to connect all the points into a forest, and then connect all the forests into a tree. Each time, the nearest two points are connected until a tree is generated. At the same time, in order to prevent the formation of a ring, we are in the process of generating a tree. Use union search to check whether a cycle will be formed after connecting these two points. If find(l)==find®, it means that a cycle must be formed after connection.

example

P3366 [Template] Minimum spanning tree

[Template] Minimum spanning tree

topic description

As in the title, given an undirected graph, find the minimum spanning tree, and output it if the graph is not connected orz.

input format

The first line contains two integers $N,M$ , indicating that the graph has a total of$N$ nodes and$M$ undirected edges.

next $M$ lines each contain three integers$X_i,Y_i,Z_i$, indicating that there is a length $Z_i$The undirected edge connects nodes $X_i,Y_i$。

output format

If the graph is connected, output an integer representing the sum of the lengths of the sides of the minimum spanning tree. Output if the graph is not connected orz.

Example #1

Sample Input #1

4 5
1 2 2
1 3 2
1 4 3
2 3 4
3 4 3

Sample output #1

7

hint

Data scale:

For $20\%$ of the data,$N\le 5$，$M\le 20$。

For $40\%$ of the data,$N\le 50$，$M\le 2500$。

For $70\%$ of the data,$N\le 500$，$M\le 10^4$。

For $100\%$ of data:$1\le N\le 5000$，$1\le M\le 2\times 10^5$，$1\le Z_i\le 10^4$。

Example explanation:

So the total edge weight of the minimum spanning tree is $2+2+3=7$。

Algorithm steps

• First sort the data according to the size of the edge
• Then traverse the edges in turn, if the two points are not connected together, then this edge is the edge of the minimum spanning tree

Algorithm implementation

struct edge
{
    
    
	int l,r,dis;
};

edge es[200010];
int fa[5010];
bool comp(edge& e1, edge& e2)
{
    
    
	return e1.dis < e2.dis;
}

void init(int n)
{
    
    
    fa[n] = n;
}
int find(int i)
{
    
    
    if (fa[i] == i)
        return i;
    fa[i] = find(fa[i]);
    return fa[i];
}
void unionn(int x, int y)
{
    
    
    int fa1 = find(x);
    int fa2 = find(y);
    fa[fa1] = fa2;
}
int main()
{
    
    
    for (int i = 0; i < 5010; i++) init(i);
	int n, m;
	cin >> n >> m;;
	for (int i = 0; i < m; i++)
		cin >> es[i].l >> es[i].r >> es[i].dis;
	sort(es, es + m, comp);
    int cnt = 0;
    for (int i = 0; i < m; i++)
    {
    
    
        int l = es[i].l;
        int r = es[i].r;
        if (find(l) == find(r))
            continue;
        unionn(l, r);
        cnt += es[i].dis;    
    }
    for(int i=1;i<n;i++)
        if (find(i) != find(i + 1))
        {
    
    
            cout << "orz" << endl;
            return 0;
        }
    cout << cnt << endl;
}