Submission #832116

#	Time	Username	Problem	Language	Result	Execution time	Memory
832116		maomao90	Split the Attractions (IOI19_split)	C++17	40 / 100	52 ms	15220 KiB

split

// I can do all things through Christ who strengthens me
// Philippians 4:13

#include "split.h"
#include "bits/stdc++.h"
using namespace std;

#define REP(i, j, k) for (int i = j; i < (k); i++)
#define RREP(i, j, k) for (int i = j; i >= (k); i--)

template <class T>
inline bool mnto(T &a, const T b) {return a > b ? a = b, 1 : 0;}
template <class T>
inline bool mxto(T &a, const T b) {return a < b ? a = b, 1 : 0;}

typedef long long ll;
typedef long double ld;
#define FI first
#define SE second
typedef pair<int, int> ii;
typedef pair<ll, ll> pll;
#define SZ(x) (int) x.size()
#define ALL(x) x.begin(), x.end()
#define pb push_back
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<ll> vll;
typedef tuple<int, int, int> iii;
typedef vector<iii> viii;

#ifndef DEBUG
#define cerr if (0) cerr
#endif

const int INF = 1000000005;
const ll LINF = 1000000000000000005;
const int MAXN = 100005;

int n, m, s[3], ss[3];
vi adj[MAXN];
vi ans;

int p[MAXN], rnk[MAXN];
void init() {
    REP (i, 0, n) {
        p[i] = i;
        rnk[i] = 1;
    }
}
int findp(int i) {
    if (p[i] == i) {
        return i;
    }
    return p[i] = findp(p[i]);
}
bool join(int a, int b) {
    int pa = findp(a), pb = findp(b);
    if (pa == pb) {
        return 0;
    }
    if (rnk[pa] < rnk[pb]) {
        swap(pa, pb);
    }
    if (rnk[pa] == rnk[pb]) {
        rnk[pa]++;
    }
    p[pb] = pa;
    return 1;
}

int sub[MAXN];
int getsub(int u, int p) {
    sub[u] = 1;
    for (int v : adj[u]) {
        if (v == p) {
            continue;
        }
        sub[u] += getsub(v, u);
    }
    return sub[u];
}
void mark(int u, int p, int &rem, int x) {
    if (rem == 0) {
        return;
    }
    ans[u] = x;
    rem--;
    for (int v : adj[u]) {
        if (v == p) {
            continue;
        }
        mark(v, u, rem, x);
    }
}

vi find_split(int n, int a, int b, int c, vi p, vi q) {
    ::n = n; s[0] = a; s[1] = b; s[2] = c;
    iota(ss, ss + 3, 0);
    sort(ss, ss + 3, [&] (int l, int r) {
            return s[l] < s[r];
            });
    m = SZ(p);
    init();
    REP (i, 0, m) {
        if (join(p[i], q[i])) {
            adj[p[i]].pb(q[i]);
            adj[q[i]].pb(p[i]);
        }
    }
    getsub(0, -1);
    ans = vi(n, ss[2] + 1);
    REP (u, 0, n) {
        for (int v : adj[u]) {
            if (sub[v] > sub[u]) {
                continue;
            }
            if (sub[v] >= s[ss[0]] && n - sub[v] >= s[ss[1]]) {
                mark(v, u, s[ss[0]], ss[0] + 1);
                mark(u, v, s[ss[1]], ss[1] + 1);
                return ans;
            }
            if (sub[v] >= s[ss[1]] && n - sub[v] >= s[ss[0]]) {
                mark(v, u, s[ss[1]], ss[1] + 1);
                mark(u, v, s[ss[0]], ss[0] + 1);
                return ans;
            }
        }
    }
    ans = vi(n, 0);
    return ans;
}

• Subtask 1: 7 / 7
• Subtask 2: 11 / 11
• Subtask 3: 22 / 22
• Subtask 4: 0 / 24
• Subtask 5: 0 / 36