Submission #395647

#	Time	Username	Problem	Language	Result	Execution time	Memory
395647		MarcoMeijer	Islands (IOI08_islands)	C++14	90 / 100	1243 ms	131076 KiB

islands

#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 1e18
#define EPS 1e-9
#define pb push_back
#define popb pop_back
#define fi first
#define se second
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 vector<T>& 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 OUT(const vector<T>& x) {RE(i,x.size()) OUT(i==0?"":" ",x[i]);}
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...); }
 
// dp
template<class T> bool ckmin(T&a, T&b) { bool bl = a > b; a = min(a,b); return bl;}
template<class T> bool ckmax(T&a, T&b) { bool bl = a < b; a = max(a,b); return bl;}
 
void program();
signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    program();
}
 
 
//===================//
//   begin program   //
//===================//
 
const int MX = 1e6+8;
const int N = (1<<20);

int n, edges;
int a[MX], b[MX], l[MX];
vi adj[MX];

// finding cycles
bitset<MX> visited;
int cycleStart;
int skippedEdge = -1;
vii cycle;
bool dfsVisit(int u, int pe) {
    if(visited[u]) {
        cycleStart = u;
        skippedEdge = pe;
        return true;
    }
    bool res = false;
    visited[u] = 1;
    FOR(e,adj[u]) {
        if(e == pe || e == skippedEdge) continue;
        if(!dfsVisit((a[e] == u ? b[e] : a[e]), e)) continue;
        
        cycle.pb({(a[e] == u ? b[e] : a[e]), l[e]});
        res = u != cycleStart;
    }
    return res;
}

// finding depth
ll curBest = 0;
lll combine(lll x, ll y) {return {max(x.fi,y), max(min(x.fi,y),x.se)};}
int i1, i2;
ll dfsDepth(int u, int pe) {
    ll res = 0;
    lll p = {0,0};
    FOR(e,adj[u]) {
        if(e == pe) continue;
        if((a[e] == u ? b[e] : a[e]) == i1) continue;
        if((a[e] == u ? b[e] : a[e]) == i2) continue;
        ll d = l[e] + dfsDepth((a[e] == u ? b[e] : a[e]),e);
        res = max(res, d);
        p = combine(p,d);
    }
    curBest = max(curBest, p.fi + p.se);
    return res;
}

void program() {
    IN(n);
    RE1(i,n) {
        IN(a[edges],l[edges]); b[edges] = i;
        adj[a[edges]].pb(edges);
        adj[b[edges]].pb(edges);
        edges++;
    }

    ll ans = 0;
    RE1(i,n) {
        if(visited[i]) continue;

        // find the cycle
        curBest = 0;
        cycle.clear();
        dfsVisit(i,-1);

        // fill a, b and c
        ll m = cycle.size(), totB = 0;
        FOR(u,cycle) totB += u.se;

        ll curB = totB;
        ll bestSum1 = -INF, bestSum2 = -INF;
        REV(i,0,m) {
            curB -= cycle[i].se;
            i1 = cycle[(i+1)%m].fi;
            i2 = cycle[(i-1+m)%m].fi;
            ll C = dfsDepth(cycle[i].fi, -1);
            curBest  = max(curBest ,        bestSum1 - curB + C);
            curBest  = max(curBest , totB + bestSum2 + curB + C);
            bestSum1 = max(bestSum1,  curB + C);
            bestSum2 = max(bestSum2, -curB + C);
        }
        ans += curBest;
    }
    OUTL(ans);
}

• Subtask 1: 33 / 33
• Subtask 2: 3 / 3
• Subtask 3: 4 / 4
• Subtask 4: 10 / 10
• Subtask 5: 10 / 10
• Subtask 6: 10 / 10
• Subtask 7: 10 / 10
• Subtask 8: 10 / 10
• Subtask 9: 0 / 10