http://poj.org/problem?id=2987
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 7869 | Accepted: 2378 |
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
题意：一个公司要开除一些员工，开除每个员工都会获得利益或者失去利益，当开除某个员工的时候，该员工的下属也必须开除；
分析：
给你两个整数n和m。
表示公司要开除n个员工，员工之间的关系有m个。
然后给你n个数字，可正可负。
表示开除这个员工所得到（正）或损失（负）的利益。
m 个关系 a b
表示b员工是a员工的下属。
闭合图定义：
它是有向图的一个点集，且这个点集的所有出边仍然指向该点集。
最大权闭合图：
每一个点有一个权值，可正可负，在所有的合法闭合图中，点权之和最大的图就是最大权闭合图。
明显，这道题目就是一个典型的最大权闭合图。
具体推理证明详见国家集训队论文：胡波涛的《最小割模型在信息学竞赛中的应用》
建图：
一个超级源点s，超级汇点t。
s连接所有点权为正的点，容量是点权。
所有点权为负的点连接汇点t，容量的点权乘以-1。
b 是 a的下属，那么连接 a b，容量无穷大。
求出最大流，那么所
有正点权的和 减去 最大流 就是最大权闭合图的最大权，就是公司的最大利益。
在残量网络中从原点s出发，一遍dfs，走还有容量的点，经过的点数就是要开除的人数。
略微证明一下：（证明转自：http://www.answeror.com/archives/27629）
1.因为不连接源汇的边的权值都是无穷大, 所以最小割一定是简单割（割边只和源点汇点有连接）, 最终被删除的点集即为去掉割以后源点s所在的集合, 记为S, 即若删除了v, 则v的所有后继都会被删除, 这就满足了原问题的唯一约束条件, 保证了解的正确性.
2.设正收益为b[i], 负收益的绝对值为c[i]. 对于任何一个负收益的顶点v[i]如果在S集合中, 说明v[i]要被删除, 此时容量c[i]被计算在切割中; 对于任何一个正收益的顶点v[i]如果在S集合中, 说明v[i]要被删除, 此时容量b[i]未被计算在切割中. 设正收益之和为B, 则B-sum{b[i]}+sum{c[j]}即为最小割的值(其中i为S中的所有正权点, j为S中的所有负权点), 等价于B-(sum{b[i]}-sum{c[j]}), 注意到括号内的值即为待求的总收益, 所以最小割对应了最大收益, 保证了解的最优性.
程序：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include"stdio.h"
#include"string.h"
#include"iostream"
#include"map"
#include"string"
#include"queue"
#include"stdlib.h"
#include"math.h"
#include"algorithm"
#include"vector"
#define M 9009
#define eps 1e-10
#define inf 1000000000000000
#define mod 1000000000
#define INF 1000000000
#define LL __int64
using namespace std;
struct node
{
LL u,v,next,w;
}edge[M*50];
LL n,t,num,head[M],dis[M],p[M],q[M],use[M],work[M];
LL mini(LL a,LL b)
{
return a<b?a:b;
}
void init()
{
t=0;
memset(head,-1,sizeof(head));
}
void add(LL u,LL v,LL w)
{
edge[t].u=u;
edge[t].v=v;
edge[t].w=w;
edge[t].next=head[u];
head[u]=t++;
edge[t].u=v;
edge[t].v=u;
edge[t].w=0;
edge[t].next=head[v];
head[v]=t++;
}
LL bfs(LL S,LL T)
{
LL rear=0;
memset(dis,-1,sizeof(dis));
q[rear++]=S;
dis[S]=0;
for(LL i=0;i<rear;i++)
{
for(LL j=head[q[i]];j!=-1;j=edge[j].next)
{
LL v=edge[j].v;
if(edge[j].w&&dis[v]==-1)
{
dis[v]=dis[q[i]]+1;
q[rear++]=v;
if(v==T)
return 1;
}
}
}
return 0;
}
LL dfs(LL S,LL a,LL T)
{
if(S==T)return a;
for(LL &i=work[S];i!=-1;i=edge[i].next)
{
LL v=edge[i].v;
if(edge[i].w&&dis[v]==dis[S]+1)
{
LL tt=dfs(v,mini(a,edge[i].w),T);
if(tt)
{
edge[i].w-=tt;
edge[i^1].w+=tt;
return tt;
}
}
}
return 0;
}
LL Dinic(LL S,LL T)
{
LL ans=0;
while(bfs(S,T))
{
memcpy(work,head,sizeof(head));
while(LL tt=dfs(S,inf,T))
ans+=tt;
}
return ans;
}
void DFS(LL u,LL f)
{
use[u]=1;
if(u!=0&&u!=n+1)
num++;
for(LL i=head[u];i!=-1;i=edge[i].next)
{
LL v=edge[i].v;
if(edge[i].w&&!use[v])
DFS(v,u);
}
}
int main()
{
LL m,i,a,b;
while(scanf("%I64d%I64d",&n,&m)!=-1)
{
LL sum=0;
for(i=1;i<=n;i++)
{
scanf("%I64d",&p[i]);
if(p[i]>0)
sum+=p[i];
}
init();
for(i=1;i<=m;i++)
{
scanf("%I64d%I64d",&a,&b);
add(a,b,inf);
}
for(i=1;i<=n;i++)
{
if(p[i]>0)
add(0,i,p[i]);
else
add(i,n+1,-p[i]);
}
LL ans=Dinic(0,n+1);
num=0;
memset(use,0,sizeof(use));
DFS(0,0);
printf("%I64d %I64d
",num,sum-ans);
}
return 0;
}