【BZOJ1391】【CEOI2008】order（最小割）

Description

有N个工作，M种机器，每种机器你可以租或者买过来. 每个工作包括若干道工序，每道工序需要某种机器来完成,你可以通过购买或租用机器来完成。 现在给出这些参数，求最大利润 。

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Solution

跟太空飞行问题类似，都是最大权闭合子图的模型。
从源点向每个工作连边，流量为利润；从每个工作向对应机器连边，流量为租用价格；从机器向回点连边，流量为机器的购买价格。答案为所有工作的利润减去最小割。

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Code

/**************************************
 * Au: Hany01
 * Prob: [BZOJ1391][CEOI2008] order
 * Date: Jul 25th, 2018
 * Email: [email protected]
**************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
typedef vector<int> VI;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = j; i < i##_end_; ++ i)
#define For(i, j ,k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define SZ(a) ((int)(a.size()))
#define ALL(a) a.begin(), a.end()
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define x first
#define y second
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define y1 wozenmezhemecaia 
#ifdef hany01
#define debug(...) fprintf(stderr, __VA_ARGS__)
#else
#define debug(...)
#endif

template<typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template<typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read() {
    register char c_; register int _, __;
    for (_ = 0, __ = 1, c_ = getchar(); !isdigit(c_); c_ = getchar()) if (c_ == '-')  __ = -1;
    for ( ; isdigit(c_); c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 2505, maxm = 3000005;

int S, T, beg[maxn], v[maxm], nex[maxm], f[maxm], e = 1, lev[maxn], cur[maxn];

inline void add(int uu, int vv, int ff, int mk = 1) {
    v[++ e] = vv, f[e] = ff, nex[e] = beg[uu], beg[uu] = e;
    if (mk) add(vv, uu, 0, 0);
}

inline int BFS() {
    static queue<int> q;
    q.push(S), Set(lev, 0), lev[S] = 1;
    while (!q.empty()) {
        int u = q.front(); q.pop();
        for (register int i = beg[u]; i; i = nex[i]) if (f[i] && !lev[v[i]])
            lev[v[i]] = lev[u] + 1, q.push(v[i]);
    }
    return lev[T];
}

int DFS(int u, int flow) {
    if (u == T) return flow;
    int res = flow, t;
    for (register int& i = cur[u]; i; i = nex[i]) if (f[i] && lev[v[i]] == lev[u] + 1) {
        f[i] -= (t = DFS(v[i], min(res, f[i]))), f[i ^ 1] += t;
        if (!(res -= t)) break;
    }
    return flow - res;
}

int main()
{
#ifdef hany01
    File("bzoj1391");
#endif

    static int n, m, Ans, tmp;

    n = read(), m = read(), T = (S = n + m + 1) + 1;
    For(i, 1, n) {
        add(S, i, tmp = read()), Ans += tmp;
        for (register int t = read(), id, w; t --; ) id = read(), w = read(), add(i, id + n, w);
    }
    For(i, 1, m) add(i + n, T, read());

    while (BFS()) Cpy(cur, beg), Ans -= DFS(S, INF);
    printf("%d\n", Ans);

    return 0;
}