审题，如果该图不连通则输出 `orz`
by Rigel @ 2024-08-16 10:17:34
没看到，谢谢
by mopeicong @ 2024-08-16 10:20:18
又交了一遍，还是不对
```cpp
#include<bits/stdc++.h>
using namespace std;
int parent[1000010],n,cnt,m;
struct node{
int from,to;
int weight;
}edge[1000010];
void add(int i,int j,int w){
cnt++;
edge[cnt].from = i;
edge[cnt].to = j;
edge[cnt].weight = w;
}
bool cmp(node x,node y){
return x.weight < y.weight;
}
int findroot(int x){
if(parent[x]!=x){
parent[x]=findroot(parent[x]);
}
return parent[x];
}
int main(){
int u,v,w,ans=0,c=0;
cin>>n>>m;
for(int i=1;i<=n;i++) parent[i]=i;
for(int i=1;i<=m;i++){
cin>>u>>v>>w;
add(u,v,w);
}
sort(edge+1,edge+1+cnt,cmp);
for(int i=1;i<=cnt;i++){
int u=edge[i].from;
int v=edge[i].to;
if(findroot(u)==findroot(v)) continue;
parent[findroot(u)] = findroot(v);
ans += edge[i].weight;
c++;
if(c==n-1){
break;
}
}
if(ans!=0) cout<<ans;
else cout<<"orz";
return 0;
}
```
后面已经加了判断了
by mopeicong @ 2024-08-16 10:22:03
```
#include<bits/stdc++.h>
using namespace std;
const int N=1e5+50,M=5e5+50;
int n,m,ans;
int p[N],sz[N];
struct Edge{
int u,v,w;
}e[M];
bool cmp(Edge x,Edge y){
return x.w<y.w;
}
int get_p(int x){
return p[x]==x?x:p[x]=get_p(p[x]);
}
void Kruskal(){
sort(e+1,e+1+m,cmp);
for(int i=1;i<=n;i++) p[i]=i,sz[i]=1;
for(int i=1;i<=m;i++){
int x=get_p(e[i].u);
int y=get_p(e[i].v);
if(x==y) continue;
p[x]=y; ans+=e[i].w;
sz[y]+=sz[x];
}
}
int main(){
cin>>n>>m;
for(int i=1;i<=m;++i)
cin>>e[i].u>>e[i].v>>e[i].w;
Kruskal();
if(sz[get_p(1)]!=n) cout<<"orz"<<endl;
else cout<<ans<<endl;
return 0;
}
```
**本蒟蒻只能帮到这了**
@[mopeicong](/user/1352625)
by apzzzx @ 2024-08-16 10:24:06
不连通 $ans$ 就一定等于 $0$ ？ 自己思考。
by Rigel @ 2024-08-16 10:25:23
@[mopeicong](/user/1352625) 你判断的不对
by Melo_DDD @ 2024-08-16 10:26:48
谢谢大家，通过了
by mopeicong @ 2024-08-16 10:36:39