分析：
圆方树板子题
每个点维护一下它子树中的颜色。
启发式合并算贡献
不过也可以分颜色用虚树做
#include<cstdio>
#include<cstring>
#include<vector>
#include<map>
#include<algorithm>
#define SF scanf
#define PF printf
#define MAXN 400010
#define MAXG 1000010
using namespace std;
typedef long long ll;
typedef map<int,int>::iterator mit;
int n,m;
int dfn[MAXN],low[MAXN],cnt;
int st[MAXN],tp;
struct node{
int u,v;node () {
}node(int u1,int v1):u(u1),v(v1) {
}
}Edge[MAXN];
int tot,n2;
vector<int> a[MAXN];
void dfs(int x,int fa=0){
dfn[x]=low[x]=++cnt;st[++tp]=x;for(int i=0;i<int(a[x].size());i++){
int u=a[x][i];if(u==fa)continue;if(dfn[u]==0){
dfs(u,x);low[x]=min(low[x],low[u]);if(low[u]>=dfn[x]){
n2++;Edge[++tot]=node(n2,x);while(1){
Edge[++tot]=node(n2,st[tp]);if(st[tp--]==u)break;}}}elselow[x]=min(low[x],dfn[u]);}
}
int idx[MAXN],col[MAXN];
map<int,int> mp[MAXN];
int totc[MAXG],tot1[MAXG],son[MAXN];
ll sum[MAXN],ans[MAXN];
void add(int x,int c,int s){
sum[x]-=1ll*mp[idx[x]][c]*(totc[c]-mp[idx[x]][c]);mp[idx[x]][c]+=s;ans[x]+=1ll*s*(totc[c]-mp[idx[x]][c]);sum[x]+=1ll*mp[idx[x]][c]*(totc[c]-mp[idx[x]][c]);
}
bool vis[MAXN];
void prepare(int x){
vis[x]=1;if(col[x])totc[col[x]]++;for(int i=0;i<int(a[x].size());i++){
int u=a[x][i];if(vis[u])continue;prepare(u); }
}
void solve(int x,int fa=0){
for(int i=0;i<int(a[x].size());i++){
int u=a[x][i];if(u==fa)continue;solve(u,x);if(son[x]==0||mp[idx[u]].size()>mp[idx[son[x]]].size())son[x]=u;}if(son[x]==0)idx[x]=++tot;else{
idx[x]=idx[son[x]];sum[x]=sum[son[x]];ans[x]+=sum[x];}if(x<=n)add(x,col[x],1);
// PF("{%d %lld %lld}\n",x,sum[x],ans[x]);for(int i=0;i<int(a[x].size());i++){
int u=a[x][i];if(u==fa||u==son[x])continue;for(map<int,int>::iterator it=mp[idx[u]].begin();it!=mp[idx[u]].end();it++)add(x,it->first,it->second);}
// PF("[%d]-\n",x);
// for(map<int,int>::iterator it=mp[idx[x]].begin();it!=mp[idx[x] ].end();it++)
// PF("<%d,%d>\n",it->first,it->second);
}
int main(){
SF("%d%d",&n,&m);for(int i=1;i<=n;i++)SF("%d",&col[i]);int u,v;for(int i=1;i<=m;i++){
SF("%d%d",&u,&v);a[u].push_back(v);a[v].push_back(u);}n2=n;for(int i=1;i<=n;i++)if(dfn[i]==0)dfs(i);for(int i=1;i<=n;i++) a[i].clear();for(int i=1;i<=tot;i++){
a[Edge[i].u].push_back(Edge[i].v);a[Edge[i].v].push_back(Edge[i].u); }tot=0;ll ansx=0;for(int i=1;i<=n2;i++)if(vis[i]==0){
prepare(i);solve(i);for(mit it=mp[idx[i]].begin();it!=mp[idx[i]].end();it++){
ansx+=1ll*tot1[it->first]*it->second;tot1[it->first]+=it->second;totc[it->first]=0;}}for(int i=1;i<=n;i++)PF("%lld\n",ans[i]+ansx);}