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\n",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;
}

标签：dep,return,int,ll,cap,NOI2006,1497,out,BZOJ
来源： https://www.cnblogs.com/yijan/p/12334442.html