poj 2987 Firing

Posted 2020-09-28 ZefengYao

Firing

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 10696		Accepted: 3226

Description

You’ve finally got mad at “the world’s most stupid” employees of yours and decided to do some firings. You’re now simply too mad to give response to questions like “Don’t you think it is an even more stupid decision to have signed them?”, yet calm enough to consider the potential profit and loss from firing a good portion of them. While getting rid of an employee will save your wage and bonus expenditure on him, termination of a contract before expiration costs you funds for compensation. If you fire an employee, you also fire all his underlings and the underlings of his underlings and those underlings’ underlings’ underlings… An employee may serve in several departments and his (direct or indirect) underlings in one department may be his boss in another department. Is your firing plan ready now?

Input

The input starts with two integers n (0 < n ≤ 5000) and m (0 ≤ m ≤ 60000) on the same line. Next follows n + m lines. The first n lines of these give the net profit/loss from firing the i-th employee individually b_i (|b_i| ≤ 10⁷, 1 ≤ i ≤ n). The remaining m lines each contain two integers i and j (1 ≤ i, j ≤ n) meaning the i-th employee has the j-th employee as his direct underling.

Output

Output two integers separated by a single space: the minimum number of employees to fire to achieve the maximum profit, and the maximum profit.

Sample Input

5 5
8
-9
-20
12
-10
1 2
2 5
1 4
3 4
4 5

Sample Output

2 2

Hint

As of the situation described by the sample input, firing employees 4 and 5 will produce a net profit of 2, which is maximum.

 

题意：公司大裁员，希望尽可能的少裁剪员工，且裁掉以后所能获得的总利润最大化，输出裁剪的人数和获得的最大利润。裁员的时候裁掉某一个人，那么连同这个人的所有下属都将会被裁掉。

根据裁员的这些性质，其实就是求最大权闭合图问题，关于最大权闭合图的介绍，可以参考：http://www.hankcs.com/wp-content/uploads/2015/01/%E7%AE%97%E6%B3%95%E5%90%88%E9%9B%86%E4%B9%8B%E3%80%8A%E6%9C%80%E5%B0%8F%E5%89%B2%E6%A8%A1%E5%9E%8B%E5%9C%A8%E4%BF%A1%E6%81%AF%E5%AD%A6%E7%AB%9E%E8%B5%9B%E4%B8%AD%E7%9A%84%E5%BA%94%E7%94%A8%E3%80%8B.pdf

最大权闭合图的求解方法是

1. 先构造网络流N，添加源点s，从s到正权值点做一条边，容量为点的权值。

2. 添加汇点t，从负权值点到t做一条边，容量为点的权值的绝对值。

3. 原来的边的容量统统设为无穷大。

4. 求解最小割，最大权=正权值之和-最小割权值

5. 残余网络中的点的个数即为裁员个数。

AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<queue>
#include<set>
#include<vector>
#include<cstring>
#include<string>
#define INF 0x3f3f3f3f3f3f3f3f
using namespace std;
const int N_MAX = 5000+4, M_MAX = 200000, V_MAX = N_MAX + 2;

typedef long long ll;
int n, m;
struct edge {
    int to; ll cap;int rev;
    edge(int to, ll cap, int rev) :to(to), cap(cap), rev(rev) {}
};
vector<edge>G[V_MAX];
int level[V_MAX];
int iter[V_MAX];

void add_edge(int from, int to, ll cap) {
    G[from].push_back(edge(to, cap, G[to].size()));
    G[to].push_back(edge(from, 0, G[from].size() - 1));
}

void bfs(int s) {
    memset(level, -1, sizeof(level));
    queue<int>que;
    level[s] = 0;
    que.push(s);
    while (!que.empty()) {
        int v = que.front(); que.pop();
        for (int i = 0; i < G[v].size(); i++) {
            edge&e = G[v][i];
            if (e.cap > 0 && level[e.to] < 0) {
                level[e.to] = level[v] + 1;
                que.push(e.to);
            }
        }
    }
}

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

ll max_flow(int s, int t) {
    ll flow = 0;
    for (;;) {
        bfs(s);
        if (level[t] < 0)return flow;
        memset(iter, 0, sizeof(iter));
        ll f;
        while ((f = dfs(s, t, INT_MAX))>0) {
            flow += f;
        }
    }
}

int res = 0;
bool vis[V_MAX];
void solve(int v) {//深搜找点
    res++;
    vis[v] = true;
    for (int i = 0; i < G[v].size();i++) {
        edge e = G[v][i];
        if (e.cap > 0 && !vis[e.to]) {
            solve(e.to);
        }
    }
}


int main() {
    while (scanf("%d%d",&n,&m)!=EOF) {
        res = 0; memset(vis, 0, sizeof(vis));
        int s = 0, t = n+1,V=t+1;
        ll sum = 0;
        for (int i = 0; i < n;i++) {
            ll a;
            scanf("%lld",&a);
            if (a > 0) { add_edge(s, i + 1, a); sum += a; }
            else add_edge(i + 1, t, -a);
       }
        for (int i = 0; i < m;i++) {
            int a, b;
            scanf("%d%d",&a,&b);
            add_edge(a,b,INF);
       }
        ll flow=max_flow(s, t);
        solve(s);
        printf("%d %lld\n",--res,sum-flow);
        for (int i = 0; i < V;i++) {//向量清空
            G[i].clear();
        }
    }
    return 0;
}