Submission #60948

#	Time	Username	Problem	Language	Result	Execution time	Memory
60948		Benq	Asceticism (JOI18_asceticism)	C++11	0 / 100	135 ms	125908 KiB

asceticism

#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 = 2001;

typedef array<ll,3> T; // distance, begin, end
T ori = {INF,INF,INF};
template<class T> void MN (T& a, T b) { a = min(a,b); }

struct dat {
    pi path;
    vector<T> besPath;
    dat() {}
    dat(pi _path, vector<T> _besPath) {
        path = _path;
        T mn = ori; for (auto a: _besPath) MN(mn,a);
        T bes[2][2]; F0R(i,2) F0R(j,2) bes[i][j] = ori;
        bes[0][0] = mn;
        for (auto a: _besPath) {
            if (a[1] == mn[1]) {
                if (a[2] != mn[2]) MN(bes[0][1],a);
            } else {
                MN(bes[1][0],a);
            }
        }
        for (auto a: _besPath) if (a[1] != mn[1] && a[2] != bes[1][0][2]) MN(bes[1][1],a);
        F0R(i,2) F0R(j,2) if (bes[i][j] != ori) besPath.pb(bes[i][j]);
    }
};

dat comb(const dat& l, const dat& r) {
    assert(l.path.s == r.path.f);
    
    vector<T> cand;
    for (auto a: l.besPath) for (auto b: r.besPath) if (a[2] != b[1]) 
        cand.pb({a[0]+b[0],a[1],b[2]});
        
    return dat({l.path.f,r.path.s},cand);
}

template<int SZ> struct Dijkstra {
    pl dist[SZ][2];
    vpi adj[SZ];
    
    void addEdge(int A, int B, int C) {
        adj[A].pb({B,C}), adj[B].pb({A,C});
    }
    
    void gen(T tmp) {
        F0R(i,SZ) F0R(j,2) dist[i][j] = {INF,INF};
        priority_queue<T,vector<T>,greater<T>> q;
        
        dist[tmp[2]][0] = {tmp[0],tmp[1]};
        q.push(tmp);
        
    	while (sz(q)) {
    		auto x = q.top(); q.pop();
    		
    		bool ok = 0;
    		F0R(j,2) if (dist[x[2]][j] == mp(x[0],x[1])) ok = 1;
    		if (!ok) continue;
    		
    		for (pi y: adj[x[2]]) {
    		    if (y.f == x[1]) continue;
    		    if (x[0]+y.s < dist[y.f][0].f) {
    		        if (dist[y.f][0].s != x[2]) dist[y.f][1] = dist[y.f][0];
        			dist[y.f][0] = {x[0]+y.s,x[2]};
        			q.push({x[0]+y.s,x[2],y.f});
    		    } else {
    		        if (x[2] == dist[y.f][0].s) continue;
    		        if (x[0]+y.s < dist[y.f][1].f) {
            			dist[y.f][1] = {x[0]+y.s,x[2]};
        			    q.push({x[0]+y.s,x[2],y.f});
    		        }
    		    }
    		}
    	}
    }
};

Dijkstra<MX> D;

int N,M,t,L;
dat d[MX][MX];

void init() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N >> M >> t >> L;
    F0R(i,M) {
        int A,B,C; cin >> A >> B >> C; 
        D.addEdge(A,B,C);
    }
}

ll get(vl X) {
    dat z = d[X[0]][X[1]];
    FOR(i,2,sz(X)) z = comb(z,d[X[i-1]][X[i]]);
    /*if (sz(X) == 3) {
        for (auto a: d[2][1].besPath) {
            cout << a[0] << " " << a[1] << " " << a[2] << "\n";
        }
        cout << "HI " << z.path.f << " " << z.path.s << "\n";
        for (auto a: z.besPath) cout << a[0] << " " << a[1] << " " << a[2] << "\n";
        exit(0);
    }*/
    ll ans = INF;
    for (auto a: z.besPath) ans = min(ans,a[0]);
    if (ans == INF) ans = -1;
    return ans;
}

int main() {
    init();
    vector<T> path[MX];
    FOR(i,1,N+1) {
        for (pi j: D.adj[i]) {
            D.gen({j.s,i,j.f});
            FOR(k,1,N+1) if (k != i)
                F0R(z,2) if (D.dist[k][z].f != INF) 
                    path[k].pb({D.dist[k][z].f,j.f,D.dist[k][z].s});
        }
        
        FOR(k,1,N+1) if (k != i) {
            /*if (k == 1) {
                for (auto a: path[k]) cout << "HI " << a[0] << " " << a[1] << " " << a[2] << "\n";
            }*/
            d[i][k] = dat({i,k},path[k]);
            /*if (k == 1) {
                for (auto a: d[i][k].besPath) cout << "hi " << a[0] << " " << a[1] << " " << a[2] << "\n";
                exit(0);
            }*/
            path[k].clear();
        }
    }
    
    vl X(L); F0R(i,L) cin >> X[i];
    F0R(i,t) {
        int p,q; cin >> p >> q;
        X[p-1] = q;
        /*FOR(j,2,sz(X)+1) {
            vl XX = vl(X.begin(),X.begin()+j);
            cout << get(XX) << "\n";
        }*/
        cout << get(X) << "\n";
    }
}

/* 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 / 4
• Subtask 2: 0 / 20
• Subtask 3: 0 / 25
• Subtask 4: 0 / 51