제출 #395696

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
395696		MarcoMeijer	Islands (IOI08_islands)	C++14	100 / 100	1093 ms	131072 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+1;
const int N = (1<<20);

int n, edges;
int l[MX];
vii adj[MX];

 
// finding cycles
bitset<MX> visited;
int cycleStart;
int skippedEdge = -1;
bool foundCycle = false;
vii cycle;
void dfsVisit(int u, int pe) {
    if(visited[u]) {
        cycleStart = u;
        skippedEdge = pe;
        foundCycle = true;
        return;
    }
    visited[u] = 1;
    RE(i,adj[u].size()) {
        if(adj[u][i].se == pe || adj[u][i].se == skippedEdge) continue;
        dfsVisit(adj[u][i].fi, adj[u][i].se^1);
        if(!foundCycle) continue;
        
        cycle.pb({adj[u][i].fi, l[adj[u][i].se/2]});
        foundCycle = u != cycleStart;
        return;
    }
}

// finding depth
ll curBest = 0;
void combine(lll& x, ll y) {x = {max(x.fi,y), max(min(x.fi,y),x.se)};}
bitset<MX> visited2;
ll dfsDepth(int u) {
    lll p = {0,0};
    visited[u] = 1;
    visited2[u] = 1;
    RE(i,adj[u].size()) {
        if(visited2[adj[u][i].fi]) continue;
        combine(p,l[adj[u][i].se/2] + dfsDepth(adj[u][i].fi));
    }
    curBest = max(curBest, p.fi + p.se);
    return p.fi;
}

void program() {
    IN(n);
    RE1(b,n) {
        int a; IN(a,l[b-1]);
        adj[a].pb({b,edges++});
        adj[b].pb({a,edges++});
    }
    RE1(i,n) adj[i].shrink_to_fit();
 
    ll ans = 0;
    RE1(i,n) {
        if(visited[i]) continue;
        
        // find the cycle
        foundCycle = false;
        curBest = 0;
        cycle.clear(); cycle.shrink_to_fit();
        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;
            visited2[cycle[i].fi] = 0;
            visited2[cycle[(i+1+m)%m].fi] = 1;
            visited2[cycle[(i-1+m)%m].fi] = 1;
            ll C = dfsDepth(cycle[i].fi);
            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: 10 / 10