結果

問題 No.1418 Sum of Sum of Subtree Size
ユーザー tada721tada721
提出日時 2021-02-05 16:21:57
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 121 ms / 2,000 ms
コード長 7,565 bytes
コンパイル時間 1,193 ms
コンパイル使用メモリ 108,828 KB
実行使用メモリ 28,416 KB
最終ジャッジ日時 2024-07-02 06:38:06
合計ジャッジ時間 5,700 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 41
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp: In function 'long long int keta(long long int)':
main.cpp:42:1: warning: control reaches end of non-void function [-Wreturn-type]
   42 | }
      | ^
main.cpp: In function 'long long int gcd(long long int, long long int)':
main.cpp:56:1: warning: control reaches end of non-void function [-Wreturn-type]
   56 | }
      | ^
main.cpp: In function 'long long int lcm(long long int, long long int)':
main.cpp:69:1: warning: control reaches end of non-void function [-Wreturn-type]
   69 | }
      | ^

ソースコード

diff # 

#include<iostream>
#include<algorithm>
#include<cmath>
#include<map>
#include<stdio.h>
#include<vector>
#include<queue>
#include<math.h>
#include<deque>
#include<set>
#include<bitset>
#include<ctime>
#include<random>
using namespace std;
#define double long double
#define int long long
#define rep(s,i,n) for(int i=s;i<n;i++)
#define c(n) cout<<n<<endl;
#define ic(n) int n;cin>>n;
#define sc(s) string s;cin>>s;
#define dc(d) double d;cin>>d;
#define mod 1000000007
#define inf 1000000000000000007
#define f first
#define s second
#define mini(c,a,b) *min_element(c+a,c+b)
#define maxi(c,a,b) *max_element(c+a,c+b)
#define pi 3.141592653589793238462643383279
#define e_ 2.718281828459045235360287471352
#define P pair<int,int>
#define upp(a,n,x) upper_bound(a,a+n,x)-a;
#define low(a,n,x) lower_bound(a,a+n,x)-a;
#define pb push_back
//printf("%.12Lf\n",);
int keta(int x) {
	rep(0, i, 30) {
		if (x < 10) {
			return i + 1;
		}
		x = x / 10;
	}
}
int gcd(int x, int y) {
	if (x == 0 || y == 0)return x + y;
	int aa = x, bb = y;
	rep(0, i, 1000) {
		aa = aa % bb;
		if (aa == 0) {
			return bb;
		}
		bb = bb % aa;
		if (bb == 0) {
			return aa;
		}
	}
}
int lcm(int x, int y) {
	int aa = x, bb = y;
	rep(0, i, 1000) {
		aa = aa % bb;
		if (aa == 0) {
			return x / bb * y;
		}
		bb = bb % aa;
		if (bb == 0) {
			return x / aa * y;
		}
	}
}
int integer(double d){
	return long(d);
}	
int distance(double a,double b,double c,double d){
	return sqrt((b-a)*(b-a)+(c-d)*(c-d));
}
bool p(int x) {
	if (x == 1)return false;
	rep(2, i, sqrt(x) + 1) {
		if (x % i == 0 && x != i) {
			return false;
		}
	}
	return true;
}
int max(int a, int b) {
	if (a >= b)return a;
	else return b;
}
string maxst(string s, string t) {
	int n = s.size();
	int m = t.size();
	if (n > m)return s;
	else if (n < m)return t;
	else {
		rep(0, i, n) {
			if (s[i] > t[i])return s;
			if (s[i] < t[i])return t;
		}
		return s;
	}
}
int min(int a, int b) {
	if (a >= b)return b;
	else return a;
}
string string_reverse(string s){
	int n=s.size();
	string t;
	rep(0,i,n)t+=s[n-i-1];
	return t;
}	
int n2[41];
int nis[41];
int nia[41];
int mody[41];
int nn;
int com(int n, int y) {
	int ni = 1;
	for (int i = 0;i < 41;i++) {
		n2[i] = ni;
		ni *= 2;
	}
	int bunsi = 1, bunbo = 1;
	rep(0, i, y)bunsi = (bunsi * (n - i)) % mod;
	rep(0, i, y)bunbo = (bunbo * (i + 1)) % mod;
	mody[0] = bunbo;
	rep(1, i, 41) {
		bunbo = (bunbo * bunbo) % mod;
		mody[i] = bunbo;
	}
	rep(0, i, 41)nis[i] = 0;
	nn = mod - 2;
	for (int i = 40;i >= 0;i -= 1) {
		if (nn > n2[i]) {
			nis[i]++;
			nn -= n2[i];
		}
	}
	nis[0]++;
	rep(0, i, 41) {
		if (nis[i] == 1) {
			bunsi = (bunsi * mody[i]) % mod;
		}
	}
	return bunsi;
}
int newcom(int n,int y){
	int bunsi = 1, bunbo = 1;
	rep(0, i, y){
		bunsi = (bunsi * (n - i)) ;
		bunbo = (bunbo * (i + 1)) ;
		int k=gcd(bunsi,bunbo);
		bunsi/=k;
		bunbo/=k;
	}	
	return bunsi/bunbo;
}	
int gyakugen(int n, int y) {
	int ni = 1;
	for (int i = 0;i < 41;i++) {
		n2[i] = ni;
		ni *= 2;
	}
	mody[0] = y;
	rep(1, i, 41) {
		y = (y * y) % mod;
		mody[i] = y;
	}
	rep(0, i, 41)nis[i] = 0;
	nn = mod - 2;
	for (int i = 40;i >= 0;i -= 1) {
		if (nn > n2[i]) {
			nis[i]++;
			nn -= n2[i];
		}
	}
	nis[0]++;
	rep(0, i, 41) {
		if (nis[i] == 1) {
			n = (n * mody[i]) % mod;
		}
	}
	return n;
}
int yakuwa(int n) {
	int sum = 0;
	rep(1, i, sqrt(n + 1)) {
		if (n % i == 0)sum += i + n / i;
		if (i * i == n)sum -= i;
	}
	return sum;
}
int poow(int y, int n) {
	if (n == 0)return 1;
	n -= 1;
	int ni = 1;
	for (int i = 0;i < 41;i++) {
		n2[i] = ni;
		ni *= 2;
	}
	int yy = y;
	mody[0] = yy;
	rep(1, i, 41) {
		yy = (yy * yy) % mod;
		mody[i] = yy;
	}
	rep(0, i, 41)nis[i] = 0;
	nn = n;
	for (int i = 40;i >= 0;i -= 1) {
		if (nn >= n2[i]) {
			nis[i]++;
			nn -= n2[i];
		}
	}
	rep(0, i, 41) {
		if (nis[i] == 1) {
			y = (y * mody[i]) % mod;
		}
	}
	return y;
}
int minpow(int x, int y) {
	int sum = 1;
	rep(0, i, y)sum *= x;
	return sum;
}
int ketawa(int x, int sinsuu) {
	int sum = 0;
	rep(0, i, 100)sum += (x % poow(sinsuu, i + 1)) / (poow(sinsuu, i));
	return sum;
}
int sankaku(int a) {
	if(a%2==0) return a /2*(a+1);
	else return (a+1)/2*a;
}
int sames(int a[1111111], int n) {
	int ans = 0;
	rep(0, i, n) {
		if (a[i] == a[i + 1]) {
			int j = i;
			while (a[j + 1] == a[i] && j <= n - 2)j++;
			ans += sankaku(j - i);
			i = j;
		}
	}
	return ans;
}
using Graph = vector<vector<int>>;
int oya[214514];
int depth[214514];
int subtreesize[214514];
void dfs(const Graph& G, int v, int p, int d) {
	depth[v] = d;
	oya[v] = p;
	for (auto nv : G[v]) {
		if (nv == p) continue; // nv が親 p だったらダメ
		dfs(G, nv, v, d + 1); // d を 1 増やして子ノードへ
	}
	subtreesize[v]=1;
	for(auto c:G[v]){
		if(c==p)continue;
		subtreesize[v]+=subtreesize[c];
	}
}
/*int H=10,W=10;
char field[10][10];
char memo[10][10];
void dfs(int h, int w) {
	memo[h][w] = 'x';

	// 八方向を探索
	for (int dh = -1; dh <= 1; ++dh) {
		for (int dw = -1; dw <= 1; ++dw) {
			if(abs(0-dh)+abs(0-dw)==2)continue;
			int nh = h + dh, nw = w + dw;

			// 場外アウトしたり、0 だったりはスルー
			if (nh < 0 || nh >= H || nw < 0 || nw >= W) continue;
			if (memo[nh][nw] == 'x') continue;

			// 再帰的に探索
			dfs(nh, nw);
		}
	}
}*/
int medi(int a,int b,int c){
	return a+b+c-max({a,b,c})-min({a,b,c});
}
int XOR(int a, int b) {
	if (a == 0 || b == 0) {
		return a + b;
	}
	int ni = 1;
	rep(0, i, 41) {
		n2[i] = ni;
		ni *= 2;
	}
	rep(0, i, 41)nis[i] = 0;
	for (int i = 40;i >= 0;i -= 1) {
		if (a >= n2[i]) {
			nis[i]++;
			a -= n2[i];
		}
		if (b >= n2[i]) {
			nis[i]++;
			b -= n2[i];
		}
	}
	int sum = 0;
	rep(0, i, 41)sum += (nis[i] % 2 * n2[i]);
	return sum;
}
//int ma[1024577][21];
//for(int bit=0;bit<(1<<n);bit++)rep(0,i,n)if(bit&(1<<i))ma[bit][i]=1;
struct UnionFind {
	vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2
 
	UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
		for (int i = 0; i < N; i++) par[i] = i;
	}
 
	int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
		if (par[x] == x) return x;
		return par[x] = root(par[x]);
	}
 
	void unite(int x, int y) { // xとyの木を併合
		int rx = root(x); //xの根をrx
		int ry = root(y); //yの根をry
		if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
		par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
	}
 
	bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
		int rx = root(x);
		int ry = root(y);
		return rx == ry;
	}
};
int n;
int a[214514],b[214514];
vector<int> ve[214514];
int solve1(){
	Graph G(n);
	rep(0,i,n-1){
		G[a[i]].pb(b[i]);
		G[b[i]].pb(a[i]);
	}
	rep(0,i,n){
		if(G[i].size()==1){
			dfs(G,i,-1,0);
			break;
		}
	}
	rep(0,i,n-1){
		if(depth[a[i]]>depth[b[i]]){
			ve[a[i]].pb(n-subtreesize[a[i]]);
			ve[b[i]].pb(subtreesize[a[i]]);
		}
		else{
			ve[b[i]].pb(n-subtreesize[b[i]]);
			ve[a[i]].pb(subtreesize[b[i]]);
		}
	}
	int ans=0;
	rep(0,i,n){
		ans+=2*n-1;
		int sum=0;
		rep(0,j,ve[i].size())sum+=ve[i][j];
		rep(0,j,ve[i].size())ans+=(sum-ve[i][j])*ve[i][j];
		ve[i].clear();
	}
	G.clear();
	return ans;
}
int solve2(){
	Graph g(n);
	rep(0,i,n-1){
		g[a[i]].pb(b[i]);
		g[b[i]].pb(a[i]);
	}
	int sum=0;
	rep(0,i,n){
		rep(0,j,n)subtreesize[j]=0;
		dfs(g,i,-1,0);
		rep(0,j,n)sum+=subtreesize[j];
	}
	g.clear();
	return sum;
}
signed main(){
	cin>>n;
	rep(0,i,n-1){
		cin>>a[i]>>b[i];
		a[i]-=1;
		b[i]-=1;
	}
	c(solve1())
}
0