Kruskal算法

空间复杂度：O(N+E)

时间复杂度：O(ElogE)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=5e3+5, maxm=2e5+5;
int f[maxn], n, m;								//f为并查集数组 

struct Edge{ int x, y, key; }edge[maxm];

inline bool cmp(Edge a, Edge b){ return a.key < b.key; }

inline int find(int x)							//并查集的路径压缩 
{
    if(f[x]==x)	return x;
    else		return f[x]=find(f[x]);
}

//inline int find(int x) { return f[x]==x ? x : f[x]=find(f[x]); }

int kruskal()
{
    sort(edge+1, edge+m+1, cmp);				//边排序 
    for(int i=1; i<=n; i++)	f[i] = i;			//每个点初始化为独立集合 
    int cnt=0, ans=0;							//ans统计加入的边数，满足mst时等于n-1 
    for(int i=1; i<=m && cnt!=n-1; i++)
    {
        int x=edge[i].x, y=edge[i].y, key=edge[i].key;
        if(find(x)!=find(y))
            ans+=key, cnt++, f[find(x)] = find(y);
    }
    return cnt==n-1 ? ans : -1;
}

int main(void)
{
    scanf("%d%d", &n, &m);
    for(int i=1; i<=m; i++)	scanf("%d%d%d", &edge[i].x, &edge[i].y, &edge[i].key);
    int ans=kruskal();
    if(ans==-1)		printf("orz
");
    else			printf("%d
", ans);
    return 0;
}