王道机试指南第11章---图论---11.3最小生成树

11.3最小生成树

Kruskal—O（ElogE）

#include<iostream>
#include<cstdio>
#include<algorithm>

using namespace std;

const int MAXN=100;

struct Edge{
    
    
	int from;
	int to;
	int length;
//	bool operator<(const Edge& e)const{	//??????????????未懂 
//	return length<e.length; 
//	}
};

//等价于 
bool cmp(Edge a,Edge b){
    
    
	return a.length<b.length;
}

Edge edge[MAXN*MAXN];		//边是顶点的两倍 
int father[MAXN];
int height[MAXN];

void Initial(int n){
    
    
	for(int i=0;i<=n;i++){
    
    
		father[i]=i;
		height[i]=0;
	}
}

int Find(int x){
    
    		//查找根节点 
	if(x!=father[x]){
    
    
		father[x]=Find(father[x]);
	}
	return father[x];
} 

void Union(int x,int y){
    
    
	x=Find(x);
	y=Find(y);
	if(x!=y){
    
    
		if(height[x]<height[y]){
    
    
			father[x]=y;
		}else if(height[y]<height[x]){
    
    
			father[y]=x;
		}else{
    
    
			father[y]=x;
			height[x]++;
		}
	}
	return;
}

int Kruskal(int n,int edgeNumber){
    
    		//返回的是权值之和,传入顶点和边的个数 
	Initial(n);
	sort(edge,edge+edgeNumber,cmp);
	int sum=0; 
	for(int i=0;i<edgeNumber;i++){
    
    
		Edge current=edge[i];
		if(Find(current.from)!=Find(current.to)){
    
    
			Union(current.from,current.to);
			sum+=current.length;
		}
	}
	return sum;	
}

int main(){
    
    
	int n;
	while(scanf("%d",&n)!=EOF){
    
    
		if(n==0){
    
    
			break;
		}
		int edgeNumber=n*(n-1)/2;
		for(int i=0;i<edgeNumber;i++){
    
    
			scanf("%d%d%d",&edge[i].from,&edge[i].to,&edge[i].length);
		}
		int answer=Kruskal(n,edgeNumber);
		printf("%d\n",answer);
	}
	
	return 0;
}

Prim-O(V²)
Prim+Heap-O(ElogV)