結果

問題 No.1507 Road Blocked
ユーザー tada721tada721
提出日時 2021-05-14 22:52:55
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 134 ms / 2,000 ms
コード長 6,314 bytes
コンパイル時間 1,271 ms
コンパイル使用メモリ 99,224 KB
実行使用メモリ 25,344 KB
最終ジャッジ日時 2024-10-02 03:19:32
合計ジャッジ時間 5,531 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge1
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 5 ms
8,064 KB
testcase_01 AC 4 ms
8,064 KB
testcase_02 AC 4 ms
8,064 KB
testcase_03 AC 93 ms
25,344 KB
testcase_04 AC 134 ms
18,688 KB
testcase_05 AC 97 ms
18,432 KB
testcase_06 AC 103 ms
18,432 KB
testcase_07 AC 104 ms
18,432 KB
testcase_08 AC 99 ms
18,432 KB
testcase_09 AC 100 ms
18,432 KB
testcase_10 AC 101 ms
18,688 KB
testcase_11 AC 102 ms
18,432 KB
testcase_12 AC 100 ms
18,432 KB
testcase_13 AC 104 ms
18,560 KB
testcase_14 AC 96 ms
18,432 KB
testcase_15 AC 95 ms
18,688 KB
testcase_16 AC 95 ms
18,432 KB
testcase_17 AC 97 ms
18,560 KB
testcase_18 AC 98 ms
18,432 KB
testcase_19 AC 104 ms
18,560 KB
testcase_20 AC 100 ms
18,560 KB
testcase_21 AC 99 ms
18,688 KB
testcase_22 AC 100 ms
18,432 KB
testcase_23 AC 101 ms
18,560 KB
testcase_24 AC 100 ms
18,432 KB
testcase_25 AC 99 ms
18,560 KB
testcase_26 AC 99 ms
18,560 KB
testcase_27 AC 99 ms
18,432 KB
testcase_28 AC 97 ms
18,432 KB
testcase_29 AC 100 ms
18,560 KB
testcase_30 AC 99 ms
18,432 KB
testcase_31 AC 99 ms
18,432 KB
testcase_32 AC 99 ms
18,432 KB
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:41:1: warning: control reaches end of non-void function [-Wreturn-type]
   41 | }
      | ^
main.cpp: In function 'long long int gcd(long long int, long long int)':
main.cpp:55:1: warning: control reaches end of non-void function [-Wreturn-type]
   55 | }
      | ^
main.cpp: In function 'long long int lcm(long long int, long long int)':
main.cpp:68:1: warning: control reaches end of non-void function [-Wreturn-type]
   68 | }
      | ^

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<tuple>
#include<string>
using namespace std;
#define ll long long
#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 mod 998244353
#define inf 2000000000000000007
#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 UF UnionFind 
#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;
		}
	}
}
bool prime(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;
}
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 n,int m){
	int pro=1;
	int nn=n;
	while(m){
		if(m%2==1)pro=pro*nn%mod;
		m=m/2;
		nn=nn*nn%mod;
	}
	return pro;
}
int poow2(int n,int m,int modulo){
	int pro=1;
	int nn=n;
	while(m){
		if(m%2==1)pro=pro*nn%modulo;
		m=m/2;
		nn=nn*nn%modulo;
	}
	return pro;
}
int inv(int n,int m){
	int t=poow(m,mod-2)%mod;
	return n*t%mod;
}
int com(int n,int m){
	if(n<m)return 0;
	int bunsi=1,bunbo=1;	
	for(int i=n-m+1;i<=n;i++)bunsi=bunsi*i%mod;
	for(int i=1;i<=m;i++)bunbo=bunbo*i%mod;
	return inv(bunsi,bunbo);
}
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) {
	return a * (a + 1) / 2;
}
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;
}
struct UnionFind {
	vector<int> par;
	UnionFind(int n):par(n){
		rep(0,i,n)par[i]=i;
	}
	int root(int x){
		if (par[x]==x)return x;
		return par[x]=root(par[x]);
	}
	void unite(int x,int y){
		int rx=root(x);
		int ry=root(y);
		if (rx==ry) return; 
		par[rx]=ry;
	}
	bool same(int x,int y){
		int rx=root(x);
		int ry=root(y);
		return rx==ry;
	}
};	
int dijkstraa[514514];
void dijkstra(int n,int m,int c[75000001],int d[75000001],int l[75000001],int siten,bool mukou){
	vector<P> e[514514];
	rep(0,i,m){
		e[c[i]].pb(P{l[i],d[i]});
		if(mukou)e[d[i]].pb(P{l[i],c[i]});
	}
	rep(0,i,n)dijkstraa[i]=inf;
	dijkstraa[siten]=0;
	priority_queue<P,vector<P>,greater<P>>pp;
	pp.push(P{0,siten});
	while(!pp.empty()){
		P t=pp.top();pp.pop();
		if(t.first!=dijkstraa[t.second])continue;
		rep(0,i,e[t.s].size()){
			P g=e[t.s][i];
			if(dijkstraa[g.second]>t.first+g.first){
				dijkstraa[g.second]=t.first+g.first;
				pp.push(P{dijkstraa[g.second],g.second});				
			}	
		}
	}
}
int dijkstra2(int shuten){
	return dijkstraa[shuten];
}
vector<int> toposo(vector<vector<int>> G,vector<int> indegree,int n){
	vector<int> sorted_vertices;
	queue<int> que;
	rep(0,i,n)if(!indegree[i])que.push(i);
	while(!que.empty()){
		int v=que.front();
		que.pop();
		rep(0,i,G[v].size()){
			int u=G[v][i];
			indegree[u]-=1;
			if(!indegree[u])que.push(u);
		}
		sorted_vertices.pb(v);
	}
	return sorted_vertices;
}
struct segtree{
	vector<int> dat;
	int n;
	segtree(int n_):n(),dat(n_*4,inf){
		int x=1;
		while(n_>=x)x*=2;
		n=x;
	}
	void update(int i,int x){
		i+=n-1;
		dat[i]=x;
		while(i>0){
			i=(i-1)/2;
			dat[i]=min(dat[i*2+1],dat[i*2+2]);
		}
	}
	int query(int a,int b){return query_sub(a,b,0,0,n);}
	int query_sub(int a,int b,int k,int l,int r){
		if(r<=a||b<=l)return inf;
		else if(a<=l&&r<=b)return dat[k];
		else{
			int vl=query_sub(a,b,k*2+1,l,(l+r)/2);
			int vr=query_sub(a,b,k*2+2,(l+r)/2,r);
			return min(vl,vr);
		}
	}
	int rightest(int a,int b,int x){return rightest_sub(a,b,x,0,0,n);}
	int rightest_sub(int a,int b,int x,int k,int l,int r){
		if(dat[k]>x||r<=a||b<=l)return a-1;
		else if(k>=n-1)return k-(n-1);
		else{
			int vr=rightest_sub(a,b,x,2*k+2,(l+r)/2,r);
			if(vr!=a-1)return vr;
			else return rightest_sub(a,b,x,2*k+1,l,(l+r)/2);
		}	
	}
	int leftest(int a,int b,int x){return leftest_sub(a,b,x,0,0,n);}
	int leftest_sub(int a,int b,int x,int k,int l,int r){
		if(dat[k]>x||r<=a||b<=l)return b;
		else if(k>=n-1)return k-(n-1);
		else{
			int vl=leftest_sub(a,b,x,2*k+1,l,(l+r)/2);
			if(vl!=b)return vl;
			else return leftest_sub(a,b,x,2*k+2,(l+r)/2,r);
		}	
	}
};
ll depth[100100],subtreesize[100100];
ll a[100100],b[100100];
vector<ll> edge[100100],v[100100];
void dfs(ll r,ll p,ll d){
	depth[r]=d;
	rep(0,i,edge[r].size())if(edge[r][i]!=p)dfs(edge[r][i],r,d+1);
	rep(0,i,edge[r].size())if(edge[r][i]!=p)subtreesize[r]+=subtreesize[edge[r][i]];	
	subtreesize[r]++;
}
signed main(){
	ic(n)
	rep(0,i,n-1){
		cin>>a[i]>>b[i];
		a[i]-=1,b[i]-=1;
		edge[a[i]].pb(b[i]);
		edge[b[i]].pb(a[i]);
	}
	dfs(0,-1,0);
	rep(0,i,n-1){
		if(depth[a[i]]>depth[b[i]]){
			v[a[i]].pb(n-subtreesize[a[i]]);
			v[b[i]].pb(subtreesize[a[i]]);
		}
		else{
			v[b[i]].pb(n-subtreesize[b[i]]);
			v[a[i]].pb(subtreesize[b[i]]);
		}
	}
	int ans=0;
	rep(0,i,n){
		rep(0,j,v[i].size()){
			ans+=v[i][j]*(n-v[i][j]);
		}
	}
	ans=(n-1)*n*(n-1)-ans;
	ans%=mod;
	c(inv(ans,(n-1)*n*(n-1)%mod))
}
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