2018.06.27Firing（最大权闭合子图）

Firing

Time Limit: 5000MS Memory Limit: 131072K

Total Submissions: 11558 Accepted: 3494

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.

Source

POJ Monthly–2006.08.27, frkstyc

首先我们要知道，这题要考察的是最大权闭合子图的姿势，不懂的OIEROIEROIER可以先看看这位大佬的博客。

学习完了最大权闭合子图的知识过后，这道题做起来应该是比较轻松的了，我们可以参照求最大权闭合子图的方法，建立源点sss和汇点ttt，根据点权的正负性分别跟源点和汇点连边，在求出最小割之后dfsdfsdfs一遍sss所在的集合就可以得出最大权闭合子图了。

代码如下：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<cstdlib>
#define inf 0x3f3f3f3f
#define N 60000
#define M 3000000
using namespace std;
inline long long read(){
	long long ans=0,w=1;
	char ch=getchar();
	while(!isdigit(ch)){
		if(ch=='-')w=-1;
		ch=getchar();
	}
	while(isdigit(ch))ans=(ans<<3)+(ans<<1)+ch-'0',ch=getchar();
	return ans*w;
}
struct Node{long long v,next,c;}e[M<<1];
long long d[N],first[N],n,m,s,t,cnt=-1,ans=0,tot=0;
bool vis[N];
inline void add(long long u,long long v,long long c){
	e[++cnt].v=v;
	e[cnt].next=first[u];
	e[cnt].c=c;
	first[u]=cnt;
	e[++cnt].v=u;
	e[cnt].next=first[v];
	e[cnt].c=0;
	first[v]=cnt;
}
inline bool bfs(){
	queue<long long>q;
	q.push(s);
	memset(d,-1,sizeof(d));
	d[s]=0;
	while(!q.empty()){
		long long x=q.front();
		q.pop();
		for(long long i=first[x];i!=-1;i=e[i].next){
			long long v=e[i].v;
			if(d[v]!=-1||e[i].c<=0)continue;
			d[v]=d[x]+1;
			if(v==t)return true;
			q.push(v);
		}
	}
	return false;
}
inline long long dfs(long long x,long long f){
	if(x==t||!f)return f;
	long long flow=f;
	for(long long i=first[x];i!=-1;i=e[i].next){
		long long v=e[i].v;
		if(d[v]==d[x]+1&&flow&&e[i].c>0){
			long long tmp=dfs(v,min(e[i].c,flow));
			if(!tmp)d[v]=-1;
			flow-=tmp;
			e[i].c-=tmp;
			e[i^1].c+=tmp;
		}
	}
	return f-flow;
}
inline void dfs1(long long p){
	vis[p]=true;
	++tot;
	for(long long i=first[p];i!=-1;i=e[i].next){
		long long v=e[i].v;
		if(e[i].c>0&&!vis[v])dfs1(v);
	}
}
int main(){
	memset(first,-1,sizeof(first));
	memset(vis,false,sizeof(vis));
	n=read(),m=read(),s=0,t=n+1;
	for(long long i=1;i<=n;++i){
		long long x=read();
		if(x>0){
			add(s,i,x);
			ans+=x;
		}
		else add(i,t,-x);
	}
	for(long long i=1;i<=m;++i){
		long long u=read(),v=read();
		add(u,v,inf);
	}
	while(bfs())ans-=dfs(s,inf);
	dfs1(s);
	printf("%lld %lld",tot-1,ans);
	return 0;
}