Educational Codeforces Round 2 E - Lomsat gelral
阿新 • • 發佈：2018-11-13
題意:每個節點有個值,求每個節點子樹眾數和
題解:可線段樹合併,維護每個數出現次數和最大出現次數,以及最大出現次數的數的和
//#pragma GCC optimize(2) //#pragma GCC optimize(3) //#pragma GCC optimize(4) //#pragma GCC optimize("unroll-loops") //#pragma comment(linker, "/stack:200000000") //#pragma GCC optimize("Ofast,no-stack-protector") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") #include<bits/stdc++.h> #define fi first #define se second #define db double #define mp make_pair #define pb push_back #define pi acos(-1.0) #define ll long long #define vi vector<int> #define mod 1000000007 #define ld long double //#define C 0.5772156649 //#define ls l,m,rt<<1 //#define rs m+1,r,rt<<1|1 #define pll pair<ll,ll> #define pil pair<int,ll> #define pli pair<ll,int> #define pii pair<int,int> //#define cd complex<double> #define ull unsigned long long //#define base 1000000000000000000 #define Max(a,b) ((a)>(b)?(a):(b)) #define Min(a,b) ((a)<(b)?(a):(b)) #define fin freopen("a.txt","r",stdin) #define fout freopen("a.txt","w",stdout) #define fio ios::sync_with_stdio(false);cin.tie(0) inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;} inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;} template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;} template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;} inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;} inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;} using namespace std; const double eps=1e-8; const ll INF=0x3f3f3f3f3f3f3f3f; const int N=100000+10,maxn=50000+10,inf=0x3f3f3f3f; vi v[N]; int root[N*22]; int ls[N*22],rs[N*22],tot,n; ll ans[N],sum[N*22],num[N*22],ma[N*22]; void pushup(int o) { ma[o]=max(ma[ls[o]],ma[rs[o]]); if(ma[ls[o]]==ma[rs[o]])sum[o]=sum[ls[o]]+sum[rs[o]]; else if(ma[ls[o]]>ma[rs[o]])sum[o]=sum[ls[o]]; else sum[o]=sum[rs[o]]; } inline int Merge(int x,int y,int l,int r) { if(l==r) { if(!x||!y) { ma[x+y]=num[x+y]=num[x]+num[y]; sum[x+y]=l; return x+y; } else { ma[x]=num[x]=num[x]+num[y]; sum[x]=l; return x; } } if(!x||!y)return x+y; int m=(l+r)>>1; ls[x]=Merge(ls[x],ls[y],l,m); rs[x]=Merge(rs[x],rs[y],m+1,r); pushup(x); return x; } void build(int &o,int pos,int l,int r) { if(!o)o=++tot; if(l==r) { ma[o]=num[o]=1; sum[o]=l; return ; } int m=(l+r)>>1; if(pos<=m)build(ls[o],pos,l,m); else build(rs[o],pos,m+1,r); pushup(o); } void debug(int o,int l,int r) { printf("%d+++%d %d %d %d %d\n",o,sum[o],num[o],ma[o],l,r); if(l==r)return ; int m=(l+r)>>1; if(ls[o])debug(ls[o],l,m); if(rs[o])debug(rs[o],m+1,r); } void dfs(int u,int f) { for(int i=0;i<v[u].size();i++) { int x=v[u][i]; if(x!=f)dfs(x,u); } for(int i=0;i<v[u].size();i++) { int x=v[u][i]; if(x!=f)root[u]=Merge(root[u],root[x],1,n); } ans[u]=sum[root[u]]; } int main() { scanf("%d",&n); for(int i=1;i<=n;i++) { int x;scanf("%d",&x); build(root[i],x,1,n); } for(int i=1;i<n;i++) { int a,b;scanf("%d%d",&a,&b); v[a].pb(b),v[b].pb(a); } dfs(1,-1); for(int i=1;i<=n;i++)printf("%lld ",ans[i]);puts(""); return 0; } /******************** 4 1 2 2 4 1 2 2 3 2 4 ********************/
也可dsu on tree,先輕重鏈剖分,每次遞迴時保留重兒子的子樹資訊,統計答案時,先遞迴輕兒子統計子樹資訊,然後統計完後刪除資訊,維護每個次數的和
//#pragma GCC optimize(2) //#pragma GCC optimize(3) //#pragma GCC optimize(4) //#pragma GCC optimize("unroll-loops") //#pragma comment(linker, "/stack:200000000") //#pragma GCC optimize("Ofast,no-stack-protector") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") #include<bits/stdc++.h> #define fi first #define se second #define db double #define mp make_pair #define pb push_back #define pi acos(-1.0) #define ll long long #define vi vector<int> #define mod 1000000007 #define ld long double //#define C 0.5772156649 //#define ls l,m,rt<<1 //#define rs m+1,r,rt<<1|1 #define pll pair<ll,ll> #define pil pair<int,ll> #define pli pair<ll,int> #define pii pair<int,int> //#define cd complex<double> #define ull unsigned long long //#define base 1000000000000000000 #define Max(a,b) ((a)>(b)?(a):(b)) #define Min(a,b) ((a)<(b)?(a):(b)) #define fin freopen("a.txt","r",stdin) #define fout freopen("a.txt","w",stdout) #define fio ios::sync_with_stdio(false);cin.tie(0) inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;} inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;} template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;} template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;} inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;} inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;} using namespace std; const double eps=1e-8; const ll INF=0x3f3f3f3f3f3f3f3f; const int N=100000+10,maxn=50000+10,inf=0x3f3f3f3f; vi v[N]; int c[N],sz[N],l[N],r[N],id[N],cnt,n; ll num[N],sum[N],maxx,ans[N]; void dfs(int u,int f) { sz[u]=1; l[u]=++cnt;id[cnt]=u; for(int i=0;i<v[u].size();i++) { int x=v[u][i]; if(x!=f) { dfs(x,u); sz[u]+=sz[x]; } } r[u]=cnt; } void solve(int u,int f,bool keep) { int ma=-1,son=-1; for(int i=0;i<v[u].size();i++) { int x=v[u][i]; if(x!=f&&ma<sz[x])ma=sz[x],son=x; } for(int i=0;i<v[u].size();i++) { int x=v[u][i]; if(x!=f&&x!=son)solve(x,u,0); } if(son!=-1)solve(son,u,1); for(int i=0;i<v[u].size();i++) { int x=v[u][i]; if(x!=f&&x!=son)for(int j=l[x];j<=r[x];j++) { sum[num[c[id[j]]]]-=c[id[j]]; num[c[id[j]]]++; sum[num[c[id[j]]]]+=c[id[j]]; maxx=max(maxx,num[c[id[j]]]); } } sum[num[c[u]]]-=c[u]; num[c[u]]++; sum[num[c[u]]]+=c[u]; maxx=max(maxx,num[c[u]]); ans[u]=sum[maxx]; if(!keep)for(int i=l[u];i<=r[u];i++) { sum[num[c[id[i]]]]-=c[id[i]]; num[c[id[i]]]--; sum[num[c[id[i]]]]+=c[id[i]]; if(sum[maxx]==0)maxx--; } } int main() { scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%d",&c[i]); sum[0]+=i; } for(int i=1;i<n;i++) { int a,b;scanf("%d%d",&a,&b); v[a].pb(b),v[b].pb(a); } dfs(1,-1); solve(1,-1,0); for(int i=1;i<=n;i++)printf("%lld ",ans[i]);puts(""); return 0; } /******************** 3 1 2 3 1 2 1 3 ********************/