Power Station POJ 4045

#include <cstdio>
#include <cstring>
#include <vector>
#include <set>
using namespace std;
const int maxn=50010;
typedef __int64 LL;
vector<int>tree[maxn];// to save the relation
LL f[maxn],g[maxn],dp[maxn];//f[u] u as root to all his sons' distance g[u]the number of u'sons
set<int> myqueue;
void dfs(int u,int pa){
/* 以1为根，所有的子树到他们的子节点的和 */
if(tree[u].size() == 1 && u != 1){
g[u]=1;
f[u]=0;
return;
}
for(int v=0;v < tree[u].size();v++){
if(tree[u][v]!=pa){
dfs(tree[u][v],u);
g[u]+=g[tree[u][v]];//the son's sons' number
f[u]+=f[tree[u][v]]+g[tree[u][v]];
}
}
g[u]++;// himself
}
void dfs2(int u,int pa){// to sum the way from his father
//再考虑从父节点来的
if(tree[u].size()==1 && u!=1){
dp[u]=dp[pa]+g[1]-(g[u]<<1);
return;
}
for(int v=0;v<tree[u].size();v++){
if(tree[u][v]!= pa){
//到某节点的距离的和=该节点子树的距离和（在dfs1中获得）+从父亲那一支子树获得的和(此时，父节点那一支看成子树)。dp[儿子]=dp[父节点]-dp[儿子]-g[儿子](儿子到父节点这条路被减了子树的子节点数的次数)    + dp[儿子]+(g[1]-g[儿子])(父树上的所有节点)
dp[tree[u][v]]=dp[u]+g[1]-(g[tree[u][v]]<<1);
dfs2(tree[u][v],u);//先算了之后再跑子树
}
}
}
int main(){
int t;
int n,I,R,a,b;
scanf("%d",&t);
LL mmin;
while(t--){
scanf("%d%d%d",&n,&I,&R);

for(int i=0;i<=n;i++){
tree[i].clear();
}
for(int i=2;i<=n;i++){
scanf("%d%d",&a,&b);
tree[a].push_back(b);
tree[b].push_back(a);
}
//        for(int i=0;i<=n;i++){
//            for(int j=0;j<tree[i].size();j++){
//                printf("%d ",tree[i][j]);
//            }
//            printf("\n");
//        }
memset(f,0,sizeof(f));
memset(g,0,sizeof(g));
dfs(1,-1);
dp[1]=f[1];
dfs2(1,-1);
mmin=dp[1];
myqueue.clear();
myqueue.insert(1);
for(int i=2;i<=n;i++){
if(dp[i]<mmin){
mmin=dp[i];
myqueue.clear();
myqueue.insert(i);
}
if(dp[i]==mmin) myqueue.insert(i);
}
//warning : output long long should be I64d
printf("%I64d\n",I*I*R*mmin);//use set not num but the op
for(set<int>::iterator it=myqueue.begin();it!=myqueue.end();++it){
printf("%d ",*it);
}
printf("\n\n");
}
}