Submission #1248324


Source Code Expand

#include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
using namespace std;

#define rep(i,a,n) for(int (i)=(a); (i)<(n); (i)++)
#define repq(i,a,n) for(int (i)=(a); (i)<=(n); (i)++)
#define repr(i,a,n) for(int (i)=(a); (i)>=(n); (i)--)
#define all(v) begin(v), end(v)
#define pb(a) push_back(a)
#define fr first
#define sc second
#define INF 2000000000
#define int long long int

#define X real()
#define Y imag()
#define EPS (1e-10)
#define EQ(a,b) (abs((a) - (b)) < EPS)
#define EQV(a,b) ( EQ((a).X, (b).X) && EQ((a).Y, (b).Y) )
#define LE(n, m) ((n) < (m) + EPS)
#define LEQ(n, m) ((n) <= (m) + EPS)
#define GE(n, m) ((n) + EPS > (m))
#define GEQ(n, m) ((n) + EPS >= (m))

typedef vector<int> VI;
typedef vector<VI> MAT;
typedef pair<int, int> pii;
typedef long long ll;

typedef complex<double> P;
typedef pair<P, P> L;
typedef pair<P, double> C;

int dx[]={1, -1, 0, 0};
int dy[]={0, 0, 1, -1};
int const MOD = 1000000007;
ll mod_pow(ll x, ll n) {return (!n)?1:(mod_pow((x*x)%MOD,n/2)*((n&1)?x:1))%MOD;}
int madd(int a, int b) {return (a + b) % MOD;}
int msub(int a, int b) {return (a - b + MOD) % MOD;}
int mmul(int a, int b) {return (a * b) % MOD;}
int minv(int a) {return mod_pow(a, MOD-2);}
int mdiv(int a, int b) {return mmul(a, minv(b));}

namespace std {
    bool operator<(const P& a, const P& b) {
        return a.X != b.X ? a.X < b.X : a.Y < b.Y;
    }
}

double memo[100010];
double ans[100010];
vector<int> G[100010];

void dfs(int idx, int par) {
    int sz = G[idx].size();
    for(auto &to : G[idx]) {
        if(to == par) {
            sz--; continue;
        }
        dfs(to, idx);
        memo[idx] += memo[to] + 1.0;
    }
    if(sz) memo[idx] /= sz;
}

void dfs2(int idx, double d_par, int par) {
    double sum = 0.0;
    int sz = G[idx].size();
    for(auto &to : G[idx]) {
        if(to == par) sum += d_par + 1.0;
        else sum += memo[to] + 1.0;
    }
    // printf("idx = %lld, d_par = %.12lf, par = %lld, sum = %.12lf\n", idx, d_par, par, sum);
    ans[idx] = (sz > 0 ? sum / sz : 0.0);
    for(auto &to : G[idx]) {
        if(to == par) continue;
        dfs2(to, sz > 1 ? (sum - memo[to] - 1.0) / (sz - 1) : 0.0, idx);
    }
}

signed main() {
    int n; cin >> n;
    rep(i,0,n-1) {
        int u, v; cin >> u >> v;
        u--; v--;
        G[u].push_back(v);
        G[v].push_back(u);
    }
    dfs(0, -1);
    dfs2(0, 0.0, -1);
    rep(i,0,n) printf("%.12f\n", ans[i]);
    return 0;
}

Submission Info

Submission Time
Task D - Driving on a Tree
User tsutaj
Language C++14 (GCC 5.4.1)
Score 410
Code Size 2883 Byte
Status RE
Exec Time 208 ms
Memory 14976 KB

Judge Result

Set Name Subtask1 Subtask2 Subtask3
Score / Max Score 190 / 190 220 / 220 0 / 390
Status
AC × 3
AC × 2
AC × 5
RE × 3
Set Name Test Cases
Subtask1 sub1_in1.txt, sub1_in2.txt, sub1_in3.txt
Subtask2 sub2_in1.txt, sub2_in2.txt
Subtask3 sub1_in1.txt, sub1_in2.txt, sub1_in3.txt, sub2_in1.txt, sub2_in2.txt, sub3_in1.txt, sub3_in2.txt, sub3_in3.txt
Case Name Status Exec Time Memory
sub1_in1.txt AC 2 ms 2560 KB
sub1_in2.txt AC 3 ms 2688 KB
sub1_in3.txt AC 161 ms 14976 KB
sub2_in1.txt AC 3 ms 2688 KB
sub2_in2.txt AC 3 ms 2688 KB
sub3_in1.txt RE 208 ms 9216 KB
sub3_in2.txt RE 171 ms 9584 KB
sub3_in3.txt RE 196 ms 13184 KB