結果

問題 No.3201 Corporate Synergy
ユーザー umimel
提出日時 2025-07-11 23:04:04
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 6,524 bytes
コンパイル時間 2,070 ms
コンパイル使用メモリ 180,364 KB
実行使用メモリ 7,848 KB
最終ジャッジ日時 2025-07-11 23:04:08
合計ジャッジ時間 2,948 ms
ジャッジサーバーID
(参考情報)
judge4 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 10 WA * 10
権限があれば一括ダウンロードができます

ソースコード

diff #

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using pll = pair<ll, ll>;
#define all(a) (a).begin(), (a).end()
#define pb push_back
#define fi first
#define se second
mt19937_64 rng(chrono::system_clock::now().time_since_epoch().count());
const ll MOD1000000007 = 1000000007;
const ll MOD998244353 = 998244353;
const ll MOD[3] = {999727999, 1070777777, 1000000007};
const ll LINF = 1LL << 60LL;
const int IINF = (1 << 30) - 1;


struct union_find{
    vector<int> par;
    vector<int> siz;
    
    union_find(int n) : par(n), siz(n, 1){
        for(int i=0; i<n; i++) par[i] = i;
    }
    
    int root(int x){
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }
 
    void unite(int x, int y){
        int rx = root(x);
        int ry = root(y);
        if (rx == ry) return;
        if (siz[rx] < siz[ry]) swap(rx, ry);
        siz[rx] += siz[ry];
        par[ry] = rx;
    }
 
    bool same(int x, int y){
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
 
    int size(int x){
        return siz[root(x)];
    }
};


template<typename T> 
struct edge{
    int from;
    int to;
    T cost;
    int id;

    edge(){}
    edge(int to, T cost=1) : from(-1), to(to), cost(cost){}
    edge(int from, int to, T cost) : from(from), to(to), cost(cost) {}
    edge(int from, int to, T cost, int id) : from(from), to(to), cost(cost), id(id){}

    void reverse(){swap(from, to);}
};

template<typename T>
struct edges : std::vector<edge<T>>{
    void sort(){
        std::sort(
            (*this).begin(),
            (*this).end(), 
            [](const edge<T>& a, const edge<T>& b){
                return a.cost < b.cost;
            }
        );
    }
};

template<typename T = bool>
struct graph : std::vector<edges<T>>{
private:
    int n = 0;
    int m = 0;
    edges<T> es;
    bool dir;

public:
    graph(int n, bool dir) : n(n), dir(dir){
        (*this).resize(n);
    }

    void add_edge(int from, int to, T cost=1){
        if(dir){
            es.push_back(edge<T>(from, to, cost, m));
            (*this)[from].push_back(edge<T>(from, to, cost, m++));
        }else{
            if(from > to) swap(from, to);
            es.push_back(edge<T>(from, to, cost, m));
            (*this)[from].push_back(edge<T>(from, to, cost, m));
            (*this)[to].push_back(edge<T>(to, from, cost, m++));
        }
    }

    int get_vnum(){
        return n;
    }

    int get_enum(){
        return m;
    }

    bool get_dir(){
        return dir;
    }

    edge<T> get_edge(int i){
        return es[i];
    }

    edges<T> get_edge_set(){
        return es;
    }
};

template<typename T>
struct redge{
    int from, to;
    T cap, cost;
    int rev;
    
    redge(int to, T cap, T cost=(T)(1)) : from(-1), to(to), cap(cap), cost(cost){}
    redge(int to, T cap, T cost, int rev) : from(-1), to(to), cap(cap), cost(cost), rev(rev){}
};

template<typename T> using Edges = vector<edge<T>>;
template<typename T> using weighted_graph = vector<Edges<T>>;
template<typename T> using tree = vector<Edges<T>>;
using unweighted_graph = vector<vector<int>>;
template<typename T> using residual_graph = vector<vector<redge<T>>>;


template<typename T>
struct dinic{
    int n;
    residual_graph<T> graph;

    dinic(residual_graph<T> &graph_){
        n = (int)graph_.size();
        graph.resize(n);

        for(int from=0; from<n; from++){
            for(redge<T> e : graph_[from]){
                graph[from].push_back(redge<T>(e.to, e.cap, e.cost, (int)graph[e.to].size()));
                graph[e.to].push_back(redge<T>(from, 0, e.cost, (int)graph[from].size()-1));
            }
        }
    }

    T max_flow(int s, int t){
        residual_graph<T> rgraph(n);
        vector<int> level(n);
        vector<int> iter(n);
        for(int from=0; from<n; from++) for(redge<T> e : graph[from]) rgraph[from].push_back(e);

        function<void()> bfs = [&](){
            for(int v=0; v<n; v++) level[v]=-1;

            queue<int> Q;
            level[s] = 0;
            Q.push(s);
            while(!Q.empty()){
                int v = Q.front();
                Q.pop();

                for(redge<T> e : rgraph[v]){
                    if(e.cap > 0 && level[e.to] < 0){
                        level[e.to] = level[v] + 1;
                        Q.push(e.to);
                    }
                }
            }
        };

        function<T(int, T)> dfs = [&](int v, T f){
            if(v == t) return f;
            for(int &i=iter[v]; i<(int)rgraph[v].size(); i++){
                redge<T> &e = rgraph[v][i];
                if(e.cap > 0 && level[v] < level[e.to]){
                    T d = dfs(e.to, min(f, e.cap));
                    if(d > 0){
                        e.cap -= d;
                        rgraph[e.to][e.rev].cap += d;
                        return d;
                    }
                }
            }

            return (T)0;
        };

        T flow = 0;
        for(;;){
            bfs();
            if(level[t] < 0) return flow;
            for(int v=0; v<n; v++) iter[v] = 0;
            T f;
            while((f = dfs(s, LINF)) > 0) flow += f;
        }
    }
};


void solve(){
    ll ans = 0LL;
    int N; cin >> N;
    vector<ll> P(N);
    for(int i=0; i<N; i++){
        cin >> P[i];
        if(0 < P[i]) ans += P[i];
    }
    int M; cin >> M;
    vector<int> U(M), V(M);
    for(int i=0; i<M; i++){
        cin >> U[i] >> V[i];
        U[i]--; V[i]--;
    }
    int K; cin >> K;
    vector<int> A(K), B(K);
    vector<ll> S(K);
    for(int i=0; i<K; i++){
        cin >> A[i] >> B[i] >> S[i];
        A[i]--; B[i]--;
        ans += S[i];
    }

    residual_graph<ll> G(N+2*K+2);
    int s = N+2*K, t = N+2*K+1;
    for(int i=0; i<N; i++){
        if(0 < P[i]) G[i].pb(redge<ll>(t, 2LL*P[i]));
        else G[s].pb(redge<ll>(i, -2LL*P[i]));
    }
    for(int i=0; i<M; i++) G[U[i]].pb(redge<ll>(V[i], LINF));
    for(int i=0; i<K; i++){
        int x = N + 2 * i;
        int y = N + 2 * i + 1;
        G[s].pb(redge<ll>(x, 0LL));
        G[x].pb(redge<ll>(A[i], S[i]));
        G[x].pb(redge<ll>(B[i], S[i]));
        G[y].pb(redge<ll>(t, 2LL*S[i]));
        G[A[i]].pb(redge<ll>(y, S[i]));
        G[B[i]].pb(redge<ll>(y, S[i]));
    }

    dinic<ll> dc(G);
    ans *= 2LL;
    ans -= dc.max_flow(s, t);
    ans /= 2LL;
    cout << ans << endl;
}

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    
    int T=1;
    //cin >> T;
    while(T--) solve();
}
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