제출 #62798

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
62798		Benq	Tug of War (BOI15_tug)	C++11	0 / 100	1117 ms	8760 KiB

tug

#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 = 60001;

int n, k, l[MX],r[MX],s[MX];
bitset<600001> b;
vi adj[MX];

int pre[MX], depth[MX];
bool visit[MX], cyc[MX];
pi oops;
int yes[2], no[2], sum;
vi v;
vpi V;

void dfs(int x) {
    visit[x] = 1; v.pb(x);
    for (auto a: adj[x]) if (a != pre[x]) {
        int t = l[a]+r[a]+n-x;
        if (visit[t]) {
            if (depth[t] < depth[x]) oops = {x,a};
        } else {
            pre[t] = a; depth[t] = depth[x]+1;
            dfs(t);
        }
    }
}

int getSum(int x) { return l[x]+r[x]+n; }

void genCyc() {
    V = {oops};
    int fin = getSum(oops.s)-oops.f;
    do {
        int nex = getSum(pre[oops.f])-oops.f;
        V.pb({nex,pre[oops.f]});
        oops.f = nex;
    } while (oops.f != fin);
}

void solve(int x) {
    v.clear(), V.clear();
    F0R(i,2) yes[i] = no[i] = 0;
    dfs(x);
    int numEdge = 0;
    for (int i: v) numEdge += sz(adj[i]);
    if (numEdge != 2*sz(v)) {
        cout << "NO";
        exit(0);
    }
    genCyc();
    for (auto i: V) cyc[i.f] = 1;
    F0R(i,sz(V)) no[i&1] += s[V[i].s];
    b = (b<<(no[0]))|(b<<(no[1]));
    // cout << "HA " << x << " " << yes[0] << " " << no[0] << " " << yes[1] << " " << no[1] << "\n";
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> n >> k;
    b[0] = 1;
    FOR(i,1,2*n+1) {
        cin >> l[i] >> r[i] >> s[i];
        adj[l[i]].pb(i), adj[r[i]+n].pb(i);
        sum += s[i];
    }
    FOR(i,1,2*n+1) if (!visit[i]) solve(i);
    // cout << "HA\n";
    F0R(i,sum+1) if (b[i] == 1 && abs(2*i-sum) <= k) {
        cout << "YES";
        exit(0);
    }
    cout << "NO";
}

/* 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: 0 / 18
• Subtask 2: 0 / 30
• Subtask 3: 0 / 23
• Subtask 4: 0 / 29