MST Unification

可以证明所有可能的边都要+1

#include<cstdio>
#include<iostream>
#include<algorithm>
const int maxn = 400100;
int fa[maxn];
int n,m;
struct T{
    int x,y,v;
    inline int operator <(const T&b)const{
        return v<b.v;
    }
}way[maxn];
inline int find(int x){ return fa[x]==x?x:fa[x]=find(fa[x]); }
int main(){
    std::ios::sync_with_stdio(false),std::cin.tie(0);
    std::cin >> n >> m;
    for(int i=1;i<=n;++i)fa[i]=i;
    for(int i=1;i<=m;++i)std::cin >> way[i].x >> way[i].y >> way[i].v;
    std::sort(way+1,way+m+1);
    int ans=0;
    for(int i=1;i<=m;++i){
        for(int j=i;j<=m && way[j].v==way[i].v;++j)ans+=find(way[j].x)!=find(way[j].y);
        for(int j=i;j<=m && way[j].v==way[i].v;++j){
            if(find(way[j].x)!=find(way[j].y))
                fa[find(way[j].x)]=find(way[j].y);
            i=j;
        }
    }
    std::cout << ans - n + 1 << '\n';
}