牛客练习赛60 旗鼓相当的对手 [长链剖分/dsu on tree]

瞎搞题,乱写都能过。

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int maxn = 2e5 + 52;
int n, k;
int a[maxn];
vector<int> g[maxn];
int son[maxn], len[maxn];
void dfs(int u, int fa) {
    for (int v : g[u]) {
        if (v == fa) continue;
        dfs(v, u);
        if (len[v] > len[son[u]]) son[u] = v;
    }
    len[u] = len[son[u]] + 1;
}
int dfn[maxn], idx = 0;
void dfsdfn(int u, int fa) {
    dfn[u] = ++idx;
    if (son[u]) dfsdfn(son[u], u);
    for (int v : g[u])
        if (v != fa && v != son[u]) dfsdfn(v, u);
}
int cnt[maxn];
int sum[maxn];
int ans[maxn];
void dfsans(int u, int fa) {
    if (son[u]) dfsans(son[u], u);
    for (int v : g[u]) {
        if (v != fa && v != son[u]) {
            dfsans(v, u);
            for (int j = 1; j <= len[v]; j++) {
                if (j == k) break;
                if(k - j < len[u]) ans[u] += cnt[dfn[u] + k - j] * sum[dfn[v] + j - 1];
                if(k - j < len[u]) ans[u] += sum[dfn[u] + k - j] * cnt[dfn[v] + j - 1];
            }
            for (int j = 1; j <= len[v]; j++) {
                cnt[dfn[u] + j] += cnt[dfn[v] + j - 1];
            }
            for (int j = 1; j <= len[v]; j++) {
                sum[dfn[u] + j] += sum[dfn[v] + j - 1];
            }
        }
    }
    cnt[dfn[u]] = 1;
    sum[dfn[u]] = a[u];
}
signed main() {
    cin >> n >> k;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 2; i <= n; i++) {
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1, 0);
    dfsdfn(1, 0);
    dfsans(1, 0);
    for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
    cout << endl;
    return 0;
}
// clang-format off
// powered by c++11
// by Isaunoya
#include<bits/stdc++.h>
#define rep(i,x,y) for(register int i=(x);i<=(y);++i)
#define Rep(i,x,y) for(register int i=(x);i>=(y);--i)
using namespace std;using db=double;using ll=long long;
using uint=unsigned int;using ull=unsigned long long;
using pii=pair<int,int>;
#define Tp template
#define fir first
#define sec second
Tp<class T>void cmax(T&x,const T&y){if(x<y)x=y;}Tp<class T>void cmin(T&x,const T&y){if(x>y)x=y;}
#define all(v) v.begin(),v.end()
#define sz(v) ((int)v.size())
#define pb emplace_back
Tp<class T>void sort(vector<T>&v){sort(all(v));}Tp<class T>void reverse(vector<T>&v){reverse(all(v));}
Tp<class T>void unique(vector<T>&v){sort(all(v)),v.erase(unique(all(v)),v.end());}inline void reverse(string&s){reverse(s.begin(),s.end());}
const int SZ=1<<23|233;
struct FILEIN{char qwq[SZ],*S=qwq,*T=qwq,ch;
#ifdef __WIN64
#define GETC getchar
#else
inline char GETC(){return(S==T)&&(T=(S=qwq)+fread(qwq,1,SZ,stdin),S==T)?EOF:*S++;}
#endif
inline FILEIN&operator>>(char&c){while(isspace(c=GETC()));return*this;}inline FILEIN&operator>>(string&s){s.clear();while(isspace(ch=GETC()));if(!~ch)return*this;s=ch;while(!isspace(ch=GETC())&&~ch)s+=ch;return*this;}
inline FILEIN&operator>>(char*str){char*cur=str;while(*cur)*cur++=0;cur=str;while(isspace(ch=GETC()));if(!~ch)return*this;*cur=ch;while(!isspace(ch=GETC())&&~ch)*++cur=ch;*++cur=0;return*this;}
Tp<class T>inline void read(T&x){bool f=0;while((ch=GETC())<48&&~ch)f^=(ch==45);x=~ch?(ch^48):0;while((ch=GETC())>47)x=x*10+(ch^48);x=f?-x:x;}
inline FILEIN&operator>>(int&x){return read(x),*this;}inline FILEIN&operator>>(ll&x){return read(x),*this;}inline FILEIN&operator>>(uint&x){return read(x),*this;}inline FILEIN&operator>>(ull&x){return read(x),*this;}
inline FILEIN&operator>>(double&x){read(x);bool f=x<0;x=f?-x:x;if(ch^'.')return*this;double d=0.1;while((ch=GETC())>47)x+=d*(ch^48),d*=.1;return x=f?-x:x,*this;}
}in;
struct FILEOUT{const static int LIMIT=1<<22;char quq[SZ],ST[233];int sz,O,pw[233];
FILEOUT(){set(7);rep(i,pw[0]=1,9)pw[i]=pw[i-1]*10;}~FILEOUT(){flush();}
inline void flush(){fwrite(quq,1,O,stdout),fflush(stdout),O=0;}
inline FILEOUT&operator<<(char c){return quq[O++]=c,*this;}inline FILEOUT&operator<<(string str){if(O>LIMIT)flush();for(char c:str)quq[O++]=c;return*this;}
inline FILEOUT&operator<<(char*str){if(O>LIMIT)flush();char*cur=str;while(*cur)quq[O++]=(*cur++);return*this;}
Tp<class T>void write(T x){if(O>LIMIT)flush();if(x<0){quq[O++]=45;x=-x;}do{ST[++sz]=x%10^48;x/=10;}while(x);while(sz)quq[O++]=ST[sz--];}
inline FILEOUT&operator<<(int x){return write(x),*this;}inline FILEOUT&operator<<(ll x){return write(x),*this;}inline FILEOUT&operator<<(uint x){return write(x),*this;}inline FILEOUT&operator<<(ull x){return write(x),*this;}
int len,lft,rig;void set(int l){len=l;}inline FILEOUT&operator<<(double x){bool f=x<0;x=f?-x:x,lft=x,rig=1.*(x-lft)*pw[len];return write(f?-lft:lft),quq[O++]='.',write(rig),*this;}
}out;
#define int long long
struct Math{
vector<int>fac,inv;int mod;
void set(int n,int Mod){fac.resize(n+1),inv.resize(n+1),mod=Mod;rep(i,fac[0]=1,n)fac[i]=fac[i-1]*i%mod;inv[n]=qpow(fac[n],mod-2);Rep(i,n-1,0)inv[i]=inv[i+1]*(i+1)%mod;}
int qpow(int x,int y){int ans=1;for(;y;y>>=1,x=x*x%mod)if(y&1)ans=ans*x%mod;return ans;}int C(int n,int m){if(n<0||m<0||n<m)return 0;return fac[n]*inv[m]%mod*inv[n-m]%mod;}
int gcd(int x,int y){return!y?x:gcd(y,x%y);}int lcm(int x,int y){return x*y/gcd(x,y);}
}math;
// clang-format on

int n, k;
const int maxn = 1e5 + 51;
int a[maxn];
vector < int > g[maxn];
int depcnt[maxn], depval[maxn];
int fa[maxn];
int dep[maxn], son[maxn], sz[maxn], ans[maxn];
void dfs(int u) {
	sz[u] = 1;
	for(int v : g[u]) {
		if(v == fa[u]) continue;
		fa[v] = u;
		dep[v] = dep[u] + 1;
		dfs(v);
		sz[u] += sz[v];
		if(sz[v] > sz[son[u]])
			son[u] = v;
	}
}

int sum = 0;
void calc(int u, int lca) {
	int d = k + dep[lca] * 2 - dep[u];
	if(d > 0) {
		sum += depval[d];
		sum += depcnt[d] * a[u];
	}
	for(int v : g[u]) 
		if(v ^ fa[u])
			calc(v , lca);
}

void add(int u, int val) {
	depcnt[dep[u]] += val;
	depval[dep[u]] += val * a[u];
	for(int v : g[u])
		if(v ^ fa[u])
			add(v , val);
}

void dfs2(int u, int kep) {
	for(int v : g[u])
		if(v ^ fa[u] && v ^ son[u])
			dfs2(v , 0);
	if(son[u])
		dfs2(son[u] , 1);
	for(int v : g[u])
		if(v ^ fa[u] && v ^ son[u])
			calc(v , u), add(v , 1);
	ans[u] = sum;
	depcnt[dep[u]] ++, depval[dep[u]] += a[u];
	if(! kep)
		add(u , -1);
	sum = 0;
}

signed main(){
	//code begin.
	in >> n >> k;
	rep(i , 1 , n)
		in >> a[i];
	rep(i , 2 , n) {
		int u , v;
		in >> u >> v;
		g[u].pb(v);
		g[v].pb(u);
	}
	dfs(1);
	dfs2(1 , 1);
	rep(i , 1 , n)
		out << ans[i] << ' ';
	return 0;
	//code end.
}
posted @ 2020-03-28 13:18  _Isaunoya  阅读(190)  评论(0编辑  收藏  举报