AcWing 859. Kruskal算法求最小生成树（稀疏图）

Posted 2022-02-19 MangataTS

题面连接

https://www.acwing.com/problem/content/861/

思路

没啥好说的，就是贪心+破圈法，然后不断将最短的边加入集合中,可以参考我这篇博客
https://acmer.blog.csdn.net/article/details/118560004

代码

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 100010

struct node
    int from,to;
    long long cost;
E[N*2];

int fa[N],n,m;

bool cmp(node a,node b)
    return a.cost<b.cost;


int find(int x)
    int k = x;
    while(k != fa[k])
        k = fa[k]; 
    while(x != fa[x]) 
        int t =fa[x];
        fa[x] = k;
        x = t;
    
    return x;


void init()
    for(int i = 1; i <= n; ++i)
    fa[i] = i;


int same(int x,int y)
    return find(x) == find(y);


void merge(int a,int b)
    a = find(a);
    b = find(b);
    if(a != b)
    fa[b] = a;

long long kruskal()
    long long ans = 0;
    sort(E+1,E+m+1,cmp);
    for(int i = 1; i <= m; ++i)
        if(same(E[i].from,E[i].to))
            continue;
        merge(E[i].from,E[i].to);
        ans += E[i].cost;
    
    return ans;//返回的是最小生成树的代价 

int main()
    int u,v,w;
    scanf("%d%d",&n,&m);
    init();
    for(int i = 1; i <= m; ++i)
        scanf("%d%d%d",&u,&v,&w);
        E[i].from = u,E[i].to = v,E[i].cost = w;
    
    long long k = kruskal();
    int loc = find(1);
    for(int i = 2; i <= n; ++i)
        if(find(i) != loc)
            k = -1;
            break;
        
    if(k == -1)
        puts("impossible");
    else
        printf("%lld\\n",k);
    return 0;