제출 #268172

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
268172		MarcoMeijer	Chase (CEOI17_chase)	C++14	40 / 100	4077 ms	49160 KiB

chase

#include <bits/stdc++.h>
using namespace std;
 
// macros
typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef pair<ll, ll> lll;
typedef tuple<int, int, int> iii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<iii> viii;
typedef vector<ll> vll;
typedef vector<lll> vlll;
#define REP(a,b,c) for(int a=int(b); a<int(c); a++)
#define RE(a,c) REP(a,0,c)
#define RE1(a,c) REP(a,1,c+1)
#define REI(a,b,c) REP(a,b,c+1)
#define REV(a,b,c) for(int a=int(c-1); a>=int(b); a--)
#define FOR(a,b) for(auto& a : b)
#define all(a) a.begin(), a.end()
#define INF 1e9
#define EPS 1e-9
#define pb push_back
#define popb pop_back
#define fi first
#define se second
#define sz size()
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
 
// input
template<class T> void IN(T& x) {cin >> x;}
template<class H, class... T> void IN(H& h, T&... t) {IN(h); IN(t...); }
 
// output
template<class T1, class T2> void OUT(const pair<T1,T2>& x);
template<class T> void OUT(const T& x) {cout << x;}
template<class H, class... T> void OUT(const H& h, const T&... t) {OUT(h); OUT(t...); }
template<class T1, class T2> void OUT(const pair<T1,T2>& x) {OUT(x.fi,' ',x.se);}
template<class... T> void OUTL(const T&... t) {OUT(t..., "\n"); }
template<class H> void OUTLS(const H& h) {OUTL(h); }
template<class H, class... T> void OUTLS(const H& h, const T&... t) {OUT(h,' '); OUTLS(t...); }
 
//===================//
//  Added libraries  //
//===================//
 
//===================//
//end added libraries//
//===================//
 
void program();
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    program();
}
 
 
//===================//
//   begin program   //
//===================//
 
const int MX = 3e5+1e4;
 
int n, m, mxbc;
ll p[MX];
vii adj[MX];
int a[MX], b[MX];
ll dp[MX][2];
int skip[MX][2];
int skipe[MX][2];
vector<pair<ll,int>> best[MX][2];
ll SM[MX];
 
ll getAns(int e, int bc);
void fillBest(int e, int bc) {
    int u = b[e];
    if(!best[u][bc].empty() && skip[u][bc] == -1) return;
    if(!best[u][bc].empty() && skip[u][bc] == a[e]) return;
 
    if(!best[u][bc].empty()) {
        best[u][bc].pb({getAns(skipe[u][bc], bc), skip[u][bc]});
        skip[u][bc] = -1;
    } else {
        FOR(v,adj[u]) if(v.fi != a[e]) {
            best[u][bc].pb({getAns(v.se, bc), v.fi});
        } else {
            skipe[u][bc] = v.se;
        }
 
        skip[u][bc] = a[e];
    }
    sort(all(best[u][bc]), greater<pair<ll,int>>());
    if(best[u][bc].sz > 2)
        best[u][bc].resize(2);
}
ll getAns(int e, int bc) {
    if(dp[e][bc] != -1) return dp[e][bc];
    int u=a[e], v=b[e];
 
    ll sm = SM[v] - ll(u == -1 ? 0 : p[u]);
    ll ans=sm;
    fillBest(e,  bc);
    FOR(p,best[v][ bc]) if(p.se != u) {
        ans = max(ans,    p.fi); // go to w without placing breadcrumpts
        break;
    }
    FOR(p,best[v][!bc]) if(p.se != u) {
        ans = max(ans, sm+p.fi); // go to w with placing breadcrumpt at v
        break;
    }
 
    return dp[e][bc]=ans;
}
 
void program() {
    IN(n,mxbc);
    RE(i,n) IN(p[i]);
    RE(i,n-1) {
        IN(a[i], b[i]); a[i]--; b[i]--;
        adj[a[i]].pb({b[i],i});
        adj[b[i]].pb({a[i],i+n-1});
    }
    RE(i,n-1) a[i+n-1] = b[i];
    RE(i,n-1) b[i+n-1] = a[i];
    m = (n-1)*2;
    RE(u,n) {
        SM[u] = 0;
        FOR(v,adj[u]) SM[u] += p[v.fi];
    }
 
    RE(j,m) dp[j][0] = 0;
    RE1(i,mxbc) {
        RE(j,m) dp[j][i%2] = -1;
        RE(j,n) best[j][i%2].clear();
        RE(j,m) getAns(j,i%2);
    }
    
    ll ans = 0;
    RE(u,n-1) {
        FOR(v,adj[u]) {
            ans = max(ans, getAns(v.se,mxbc%2));
            ans = max(ans, SM[u]+getAns(v.se,(mxbc+1)%2));
        }
    }
    if(mxbc == 0) ans = 0;
    OUTL(ans);
}

• Subtask 1: 20 / 20
• Subtask 2: 20 / 20
• Subtask 3: 0 / 30
• Subtask 4: 0 / 30