HDU 6133 Army Formations(线段树合并)

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#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<vector>using namespace std;const int MAXN=(int)1e5+7;typedef long long ll;int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}struct edge{int to,nxt;}ed[MAXN<<1];int head[MAXN],cnt;void addedge(int u,int v){ed[cnt].to=v;ed[cnt].nxt=head[u];head[u]=cnt++;}int root[MAXN],a[MAXN];int ls[MAXN*18],rs[MAXN*18],tol;int num[MAXN*18];ll sum[MAXN*18],ans[MAXN*18];int n,vsz;vector<int>ve;void insert(int &rt,int l,int r,int x){    if(rt==0)rt=++tol;        ans[rt]=sum[rt]=ve[x-1];num[rt]=1;    if(l==r)return ;    int mid=l+r>>1;    if(x<=mid)insert(ls[rt],l,mid,x);    else insert(rs[rt],mid+1,r,x);}int merge(int u,int v){    if(u==0||v==0)return u|v;    ls[u]=merge(ls[u],ls[v]);    rs[u]=merge(rs[u],rs[v]);    sum[u]+=sum[v];    num[u]+=num[v];    if(ls[u]==0&&rs[u]==0)ans[u]=(sum[u]/num[u]+sum[u])*num[u]/2;else ans[u]=ans[ls[u]]+ans[rs[u]]+num[rs[u]]*sum[ls[u]];    return u;}void dfs(int u,int pre){insert(root[u],1,vsz,a[u]);for(int i=head[u];i!=-1;i=ed[i].nxt){int v=ed[i].to;if(v!=pre){dfs(v,u);merge(root[u],root[v]);}}}int main(){int t;t=read();while(t--){ve.clear();n=read();memset(head,-1,sizeof(head));memset(root,0,sizeof(root));for(int i=1;i<=n;i++)a[i]=read(),ve.push_back(a[i]);sort(ve.begin(),ve.end());ve.erase(unique(ve.begin(),ve.end()),ve.end());for(int i=1;i<=n;i++)a[i]=lower_bound(ve.begin(),ve.end(),a[i])-ve.begin()+1;vsz=ve.size();for(int i=1;i<n;i++){int u,v;u=read();v=read();addedge(u,v);addedge(v,u);}dfs(1,0);for(int i=1;i<=n;i++)printf("%lld ",ans[root[i]]);printf("\n");for(int i=0;i<=tol;i++)ls[i]=rs[i]=num[i]=ans[i]=sum[i]=0;tol=cnt=0;}return 0;}
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