POJ 2987 Firing(最大權閉合圖)

Firing

Time Limit: 5000MS	Memory Limit: 131072K
Total Submissions:11472	Accepted: 3468

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 bi (|bi| ≤ 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.

題解：
比較裸的最大權閉合圖，建立一個源點，與所有正點相連，邊權為其正值
建立一個彙點，與所有負點相連，邊權為其絕對值，之前的邊改為無窮，
即求源點到彙點的最大流，從起點出發周遊e.cap>0的點便是路徑
代碼：

#include<cstdio>
#include<string.h>
#include<vector>
#include<queue>
using namespace std;
#define ll long long
const int maxn = 5005;
const ll inf = 1e18;
struct node
{
	int to, rev; ll cap;
};
vector<node>G[maxn];
int n, m, level[maxn], iter[maxn];
bool vis[maxn];
void add(int from, int to, ll cap)
{
	node e; e.to = to; e.cap = cap; e.rev = G[to].size();
	G[from].push_back(e);
	e.to = from; e.cap = 0; e.rev = G[from].size() - 1;
	G[to].push_back(e);
}
void bfs()
{
	queue<int>P;
	memset(level, -1, sizeof(level));
	level[0] = 0; P.push(0);
	while (!P.empty())
	{
		int v = P.front(); P.pop();
		for (int i = 0; i<G[v].size(); i++)
		{
			node e = G[v][i];
			if (e.cap>0 && level[e.to]<0)
			{
				level[e.to] = level[v] + 1;
				P.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++)
	{
		node &e = G[v][i];
		if (e.cap>0 && level[e.to]>level[v])
		{
			int d;
			if (e.cap>f)d = dfs(e.to, t, f);
			else d = dfs(e.to, t, e.cap);
			if (d>0)
			{
				e.cap -= d;
				G[e.to][e.rev].cap += d;
				return d;
			}
		}
	}
	return 0;
}
ll max_flow()
{
	ll flow = 0;
	while (1)
	{
		bfs();
		if (level[n + 1]<0)return flow;
		memset(iter, 0, sizeof(iter));
		int f;
		while ((f = dfs(0, n + 1, inf))>0)flow += f;
	}
}
int upstream(int s)
{
    int cnt=0;
    memset(vis,0,sizeof(vis));
    queue<int>P;vis[s]=1;P.push(s);
    while(!P.empty())
    {
        int u=P.front();P.pop();
        for(int i=0;i<G[u].size();i++)
        {
            node e=G[u][i];
            if(e.cap>0&&!vis[e.to])
            {
                vis[e.to]=1;
                P.push(e.to);
                cnt++;
            }
        }
    }
    return cnt;
}
int main()
{
	scanf("%d%d", &n, &m);
	ll sum=0;
	for (int i = 1; i <= n; i++)
	{
		int x; scanf("%d", &x);
		if (x >= 0)add(0, i, x);
		else add(i, n + 1, -x);
		if(x>0)sum+=x;
	}
	for (int i = 1; i <= m; i++)
	{
		int x, y; scanf("%d%d", &x, &y);
		add(x, y, inf);
	}
	printf("%d %lld\n",upstream(0),sum-max_flow());
	for(int i=1;i<=n;i++)if(vis[i])printf("%d\n",i);
	return 0;
}