Submission #31016

#	Time	Username	Problem	Language	Result	Execution time	Memory
31016		kajebiii	Towns (IOI15_towns)	C++14	35 / 100	33 ms	2076 KiB

towns

#include "towns.h"
#include <bits/stdc++.h>

using namespace std;

#define SZ(v) ((int)(v).size())
#define ALL(v) (v).begin(),(v).end()
#define one first
#define two second
typedef long long ll;
typedef unsigned long long ull;
typedef pair<double, double> pd;
typedef pair<int, int> pi;
typedef pair<ll, int> pli;
typedef pair<ll, ll> pll; typedef pair<ll, pi> plp;
typedef tuple<int, int, int> ti; typedef tuple<ll, int, int> tli;
const int INF = 0x3f2f1f0f;
const ll LINF = 1ll * INF * INF * 2;

map<pi, int> Mp;
int getD(int i, int j) {
    if(Mp.count(pi(i, j)) == 0) Mp[pi(i, j)] = Mp[pi(j, i)] = getDistance(i, j);
    return Mp[pi(i, j)];
}

int N, Sub;
int hubDistance(int n, int sub) {
    Mp.clear();
    N = n; Sub = sub;

    int max0D = -1, f = -1;
    for(int i=0; i<N; i++) {
        if(i == 0) continue;
        int d = getD(0, i);
        if(max0D < d) max0D = d, f = i;
    }
    int maxfD = -1, s = -1;
    for(int i=0; i<N; i++) {
        if(i == f) continue;
        int d = getD(f, i);
        if(maxfD < d) maxfD = d, s = i;
    }

    int R = INF;
    /*
    int from0 = (getD(0, f) + getD(0, s) - getD(s, f)) / 2;
    for(int i=0; i<N; i++) {
        int dis = (getD(0, i) + getD(0, f) - getD(i, f)) / 2;
        if(dis >= from0) {
            dis = max0D - dis;
            R = min(R, max(dis, maxfD - dis));
        }
    }
    */
    for(int i=0; i<N; i++) {
        int dis = (getD(f, i) - getD(i, s) + getD(s, f)) / 2;
        R = min(R, max(dis, getD(s, f) - dis));
    }
    vector<pi> val;
    for(int i=0; i<N; i++) {
        int dis = (getD(f, i) - getD(i, s) + getD(s, f)) / 2;
        val.push_back(pi(dis, i));
    }
    sort(ALL(val));
    for(int i=0, j=0; i<N; i=j) {
        while(j < N && val[i].one == val[j].one) j++;
        int dis = val[i].one;
//        printf("[%d %d]\n", i, j);
        if(R == max(dis, getD(s, f) - dis)) {
            if(j - i <= (N/2) && i <= (N/2) && (N-j) <= (N/2)) return R;
        }
    }

    return -R;
}

• Subtask 1: 13 / 13
• Subtask 2: 12 / 12
• Subtask 3: 0 / 13
• Subtask 4: 10 / 10
• Subtask 5: 0 / 13
• Subtask 6: 0 / 39