제출 #1291126

#제출 시각아이디문제언어결과실행 시간메모리
1291126LOLOLOTug of War (BOI15_tug)C++17
100 / 100
1361 ms11772 KiB
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
 
#define           f     first
#define           s     second
#define           pb    push_back
#define           ep    emplace
#define           eb    emplace_back
#define           lb    lower_bound
#define           ub    upper_bound
#define       all(x)    x.begin(), x.end()
#define      rall(x)    x.rbegin(), x.rend()
#define   uniquev(v)    sort(all(v)), (v).resize(unique(all(v)) - (v).begin())
#define     mem(f,x)    memset(f , x , sizeof(f))
#define        sz(x)    (ll)(x).size()
#define  __lcm(a, b)    (1ll * ((a) / __gcd((a), (b))) * (b))
#define          mxx    *max_element
#define          mnn    *min_element
#define    cntbit(x)    __builtin_popcountll(x)
#define       len(x)    (int)(x.length())
 
const int N = 1e5 + 100;
set <pair <int, int>> ed[N];
bool used[N];
int st = 0, s[N];

pair <int, int> dfs(int u, int v) {
    if (used[u])
        return {0, 0};

    used[u] = 1; 
    pair <int, int> A = {0, 0};
    for (auto x : ed[u]) {
        if (x.s == v)
            continue;
 
        A = dfs(x.f, x.s);
        A.f += s[x.s];
        swap(A.f, A.s);
    }
 
    return A;
}
 
int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
 
    int n, k;
    cin >> n >> k;
 
    int S = 0;
    for (int i = 1; i <= n * 2; i++) {
        int l, r;
        cin >> l >> r >> s[i];
        ed[l].insert({r + n, i});
        ed[r + n].insert({l, i});
        S += s[i];
    } 
 
    queue <int> pq;
    int V1 = 0, V2 = 0;
    int sz = n * 2;
    for (int i = 1; i <= sz; i++) {
        if (sz(ed[i]) == 0) {
            cout << "NO\n";
            return 0;
        }
 
        if (sz(ed[i]) == 1)
            pq.push(i);
    }
 
    int L = 0;
    while (sz(pq)) {
        int t = pq.front();
        pq.pop();

        if (sz(ed[t]) == 0) {
            continue;
        }
 
        auto it = *ed[t].begin();
        int nxt = it.f, cost = it.s;
        pair <int, int> pat = {t, cost};
 
        ed[nxt].erase(pat);
        ed[t].clear();
        if (t <= n) {
            L += s[cost];
            V1++;
        } else {
            V2++;
        }
 
        if (sz(ed[nxt]) == 1) {
            pq.push(nxt);
        }
    }

    assert(V1 == V2);
 
    for (int i = 1; i <= sz; i++) {
        if (sz(ed[i]) > 2) {
            cout << "NO\n";
            return 0;
        }
    }
 
    bitset <1000001> dp;
    dp[0] = 1;
 
    for (int i = 1; i <= sz; i++) {
        if (sz(ed[i]) == 2 && used[i] == 0) {
            auto it = *ed[i].begin();
            pair <int, int> A = dfs(i, it.s);
            dp = (dp << A.f) | (dp << A.s);
        }
    }
 
 
    for (int i = 0; i <= 600000; i++) {
        if (dp[i]) {
            if (abs((S - (L + i)) - (L + i)) <= k) {
                cout << "YES\n";
                return 0;
            }
        }
    }
 
    cout << "NO\n";
    return 0;
}
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