"Minimum Spanning Tree Kruscal Algorithm, Union Check Code"
Foreword: This is the first time to write an algorithm blog hhh without the sample questions. After reading a lot of information on the Internet, I summarized it in the blog.
1. The idea of Kruscal algorithm:
① When we do the minimum spanning tree problem, we need to use an edge structure array:
typedef struct stu ///边结构体
{
int u; //边起始节点
int v; //边结尾节点
int len; //边权值
}edge;
edge a[10005]; //边的结构体数组
②Kruskal's algorithm selects the edges with the smallest weight from all the edges in turn and adds them until the added edge is equal to the number of nodes -1 (of course, if the newly added edge forms a ring, it is discarded)
③ So we need to sort the array from small to large first. There may be a small problem at this time. How to sort the structure, then we need to use C++ sort and pseudo-functions to achieve it. First, we need to define a pseudo-function compareEdge :
bool compareEdge(edge a1,edge a2) ///伪函数
{
return a1.len < a2.len;
}
Then use sort():
sort(&a[1],&a[n+1],compareEdge); ///排序边权
(If you are not familiar with STL and pseudo-functions, go to make up the lesson yourself)
④Now we start traversing the array from a[1], we have to consider whether it will form a ring every time we select an edge, usually we need to use the combined search to judge , Let’s explain the idea of combining and checking:
※Here is an analogy : now there are 3 people, 2 is 1 colleague, and 3 is 1 colleague.
So how do we judge whether these three people are in a colleague relationship? We use an array person[i] to indicate who is the highest-level boss of the i-th person. At the beginning, everyone’s boss is its own person. [i] = i , for example, 2 is a colleague of 1, then person[1] = 2, which is equivalent to 2 is now the boss of 1. 3 is a colleague of 1, then person[1] should be equal to 3 again. To avoid this phenomenon, you must find the boss of 1 first, and then **update the "boss"** of the boss of 1 to realize the link of the relationship (I It feels like the idea of a linked list), the current boss of 1 is 2, then find person[2], the original person[2] is itself, now we need to update it to 3, then the structure we imagined is like this: the
person array should It's like this:
How do we find someone’s boss?
int findHead(int x)
{
while(person[x] != x) //只要他的boss不是他自己
x = person[x]; //x等于他的下一级
return x; //最后return -> boss的值
}
When we learn to think, we will find that 1 and 2 are colleagues, and 1 and 3 are colleagues. Then 2 and 3 must be colleagues. The values of person[2] and person[3] are the same. If you add A 2 and 3 are colleagues (edges), then a ring is formed, so when the boss of au is equal to the boss of av, a ring will appear:
int judge(int x) //判断每次遍历的最小边
{
int father;
int father1 = findHead(a[x].u); //找Boss1
int father2 = findHead(a[x].v); //找Boss2
if(father1 == father2){
//相等
cout << "第"<<x<<"次有环"<<endl; //有环
return 0;
}
else
{
father = findHead(a[x].u);
person[father] = a[x].v; //没有环就更新当前点的boss
ans += a[x].len; //最小生成树+=len
cout << "第"<<x<<"次无环"<<endl;
return 1;
}
}
Finally, it is judged that the number of edges that have been added is equal to the number of nodes-1 , and the minimum spanning tree is completed:
for(int i = 1;i<=n;i++)
{
if(judge(i)) num++; //如果判断没有环，边数++
if(num == n - 1) break;
}
Finally, the complete code:
#include<bits/stdc++.h>
using namespace std;
typedef struct stu ///边结构体
{
int u;
int v;
int len;
}edge;
bool compareEdge(edge a1,edge a2) ///伪函数
{
return a1.len < a2.len;
}
int person[10005];
edge a[10005];
int ans = 0;
int findHead(int x)
{
while(person[x] != x)
x = person[x];
return x;
}
int judge(int x)
{
int father;
int father1 = findHead(a[x].u);
int father2 = findHead(a[x].v);
if(father1 == father2){
cout << "第"<<x<<"次有环"<<endl;
return 0;
}
else
{
father = findHead(a[x].u);
person[father] = a[x].v;
ans += a[x].len;
cout << "第"<<x<<"次无环"<<endl;
return 1;
}
}
int main(){
int n,m;
int num = 0;
cin >> n >> m;
for(int i=1;i<=10005;i++)
{
person[i] = i; ///初始化person数组
}
for(int i=1;i<=n;i++)
{
cin >>a[i].u >>a[i].v>>a[i].len; ///输入边数据
}
sort(&a[1],&a[n+1],compareEdge); ///排序边权数组
for(int i = 1;i<=m;i++)
{
if(judge(i)) num++;
if(num == n - 1) break;
}
cout << "最小生成树权重为："<<ans;
return 0;
}
Test a use case:
