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  • BZOJ 1497 [NOI2006]最大获利

    BZOJ 1497 [NOI2006]最大获利

    从原点向每个中转站建立容量为 $ p_i $ 的边,再从中转站向以它作为条件的用户群建立 $ +infin $ 的边,最后从用户群向汇点建立 $ c_i $ 的边,医院的权值和 - 最小割就是答案。

    为什么这样是对的呢?考虑我们满足一个用户组的需求,无非就是不要这个需求(割掉这个边)或者花代价去割掉它的 $ A_i,B_i $ ,这样做它的盈利是 $ c_i - p_a-p_b $ 。所以最后求和 c 减去最小割就是答案。

    #include "iostream"
    #include "algorithm"
    #include "cstring"
    #include "cstdio"
    #include "queue"
    #include "cmath"
    #include "vector"
    using namespace std;
    #define mem(a) memset( a , 0 , sizeof a )
    #define MAXN 6006
    #define inf 0x3f3f3f3f
    
    typedef long long ll;
    class maxFlow {
    private:
        int add(int u, int v, ll w) {
            nxt.push_back(head[u]);
            int x = ( head[u] = to.size() );
            to.push_back(v);
            cap.push_back(w);
            return x;
        }
    public:
        std::queue<int> q;
        std::vector<int> head, cur, nxt, to, dep;
        std::vector<ll> cap;
    
        maxFlow(int _n = 0) { init(_n); }
        void init(int _n) {
            head.clear();
            head.resize(_n + 1, 0);
            nxt.resize(2);
            to.resize(2);
            cap.resize(2);
        }
        void init() { init(head.size() - 1); }
        int Add(int u, int v, ll w) {
    //      printf("%d %d %d
    ",u,v,w);
            add(u, v, w);
            return add(v, u, 0);
        }
        void del(int x) { cap[x << 1] = cap[x << 1 | 1] = 0; }
        bool bfs(int s, int t, int delta) {
            dep.clear();
            dep.resize(head.size(), -1);
            dep[s] = 0;
            q.push(s);
            while (!q.empty()) {
                int u = q.front();
                q.pop();
                for (int i = head[u]; i; i = nxt[i]) {
                    int v = to[i];
                    ll w = cap[i];
                    if (w >= delta && dep[v] == -1) {
                        dep[v] = dep[u] + 1;
                        q.push(v);
                    }
                }
            }
            return ~dep[t];
        }
        ll dfs(int u, ll flow, int t, int delta) {
            if (dep[u] == dep[t])
                return u == t ? flow : 0;
            ll out = 0;
            for (int& i = cur[u]; i; i = nxt[i]) {
                int v = to[i];
                ll w = cap[i];
                if (w >= delta && dep[v] == dep[u] + 1) {
                    ll f = dfs(v, std::min(w, flow - out), t, delta);
                    cap[i] -= f;
                    cap[i ^ 1] += f;
                    out += f;
                    if (out == flow)
                        return out;
                }
            }
            return out;
        }
        ll maxflow(int s, int t) {
            ll out = 0;
            ll maxcap = *max_element(cap.begin(), cap.end());
            for (ll delta = 1ll << int(log2(maxcap) + 1e-12); delta; delta >>= 1) {
                while (bfs(s, t, delta)) {
                    cur = head;
                    out += dfs(s, 0x7fffffffffffffffll, t, delta);
                }
            }
            return out;
        }
        ll getflow(int x) const { return cap[x << 1 | 1]; }
    } F ;
    int n , m , s = 60001 , t = 60002;
    int p[MAXN];
    long long S;
    int main() {
        cin >> n >> m;
        F.init( 60006 );
        for( int i = 1 ; i <= n ; ++ i ) scanf("%d",&p[i]) , F.Add( s , i , p[i] );
        for( int i = 1,a,b,c ; i <= m ; ++ i ) {
            scanf("%d%d%d",&a,&b,&c);
            F.Add( i + n , t , c ) , F.Add( a , i + n , inf ) , F.Add( b , i + n , inf );
            S += c;
        }
        cout << S - F.maxflow( s , t ) << endl;
    }
    
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  • 原文地址:https://www.cnblogs.com/yijan/p/12334442.html
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