Submission #59752

#	Time	Username	Problem	Language	Result	Execution time	Memory
59752		Benq	File Paths (BOI15_fil)	C++11	100 / 100	325 ms	27132 KiB

fil

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
 
using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
 
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
 
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
 
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
 
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
 
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
 
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 6001;
 
int n,m,k,s,p[MX],l[MX];
ll depth[MX];
vi child[MX];
bitset<MX> ans;
bitset<1000001> root;
 
void dfs0(int x, ll cdepth) {
    cdepth += l[x];
    depth[x] = cdepth;
    for (int i: child[x]) dfs0(i,cdepth);
}
 
void genCyc(int x, int ori, ll cdepth, vi& CYC) {
    if (x > n) return;
    if (cdepth <= k) CYC.pb(cdepth);
    for (int i: child[x]) genCyc(i,ori,cdepth+l[i],CYC);
}

int cyc[1000001];

void DFS(int cur) {
    if (cur > n) {
        ll need = k-depth[cur];
        if (need == 0) {
            ans[cur] = 1;
            return;
        }
        for (int t = 1; t*t <= need; t++) if (need%t == 0) 
            if (cyc[t] || cyc[need/t]) ans[cur] = 1;
    }
    vi CYC; genCyc(cur,cur,s,CYC);
    for (int i: CYC) cyc[i] ++;
    for (int i: child[cur]) DFS(i);
    for (int i: CYC) cyc[i] --;
}
 
void case0() {
    dfs0(0,0);
    DFS(0);
}
 
void case1() {
    F0R(i,n+1) if (depth[i]+s <= k) root[depth[i]+s] = 1;
 
    FOR(i,n+1,n+m+1) {
        vi v = {l[i]};
        int x = p[i];
        while (x) {
            v.pb(v.back()+l[x]);
            x = p[x];
        }
        for (int x: v) if (x <= k && root[k-x]) ans[i] = 1;
    }
}
 
int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> n >> m >> k >> s; s ++;
    FOR(i,1,n+1) {
        cin >> p[i] >> l[i]; l[i] ++;
        child[p[i]].pb(i);
    }
    FOR(i,n+1,n+m+1) {
        cin >> p[i] >> l[i]; l[i] ++;
        child[p[i]].pb(i);
    }
    case0();
    case1();
    FOR(i,n+1,n+m+1) {
        if (ans[i]) cout << "YES";
        else cout << "NO";
        cout << "\n";
    }
}
 
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

• Subtask 1: 33 / 33
• Subtask 2: 33 / 33
• Subtask 3: 34 / 34