| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 7126 | Accepted: 2164 |
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
----------
最大权=正权和-最小割
从原点开始对残留网络dfs，能访问到的点裁去。
-----------
/*==============================================*
| Code Library
*==============================================*/
/** head-file **/
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <algorithm>
/** define-for **/
#define REP(i, n) for (int i=0;i<int(n);++i)
#define FOR(i, a, b) for (int i=int(a);i<int(b);++i)
#define DWN(i, b, a) for (int i=int(b-1);i>=int(a);--i)
#define REP_1(i, n) for (int i=1;i<=int(n);++i)
#define FOR_1(i, a, b) for (int i=int(a);i<=int(b);++i)
#define DWN_1(i, b, a) for (int i=int(b);i>=int(a);--i)
#define REP_N(i, n) for (i=0;i<int(n);++i)
#define FOR_N(i, a, b) for (i=int(a);i<int(b);++i)
#define DWN_N(i, b, a) for (i=int(b-1);i>=int(a);--i)
#define REP_1_N(i, n) for (i=1;i<=int(n);++i)
#define FOR_1_N(i, a, b) for (i=int(a);i<=int(b);++i)
#define DWN_1_N(i, b, a) for (i=int(b);i>=int(a);--i)
/** define-useful **/
#define clr(x,a) memset(x,a,sizeof(x))
#define sz(x) int(x.size())
#define see(x) cerr<<#x<<" "<<x<<endl
#define se(x) cerr<<" "<<x
#define pb push_back
#define mp make_pair
/** test **/
#define Display(A, n, m) {
REP(i, n){
REP(j, m) cout << A[i][j] << " ";
cout << endl;
}
}
#define Display_1(A, n, m) {
REP_1(i, n){
REP_1(j, m) cout << A[i][j] << " ";
cout << endl;
}
}
using namespace std;
/** typedef **/
typedef long long LL;
/** Add - On **/
const int direct4[4][2]={ {0,1},{1,0},{0,-1},{-1,0} };
const int direct8[8][2]={ {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int direct3[6][3]={ {1,0,0},{0,1,0},{0,0,1},{-1,0,0},{0,-1,0},{0,0,-1} };
const int MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long INFF = 1LL << 60;
const double EPS = 1e-9;
const double OO = 1e15;
const double PI = acos(-1.0); //M_PI;
const int maxn=11111;
const int maxm=1111111;
bool vis[maxn];
int cnt;
/*==============================================*
| Dinic最大流
| INIT: prepare(n,S,T);addedge(u,v,c);节点0~n
| CALL: Dinic_flow():int
*==============================================*/
struct edgenode{
int to;
LL flow;
int next;
};
struct Dinic{
int node,src,dest,edge;
int head[maxn],work[maxn],dis[maxn],q[maxn];
edgenode edges[maxm];
void prepare(int _node,int _src,int _dest){
node=_node,src=_src,dest=_dest;
memset(head,-1,sizeof(head));
edge=0;
}
void addedge(int u,int v,LL c){
edges[edge].flow=c,edges[edge].to=v,edges[edge].next=head[u],head[u]=edge++;
edges[edge].flow=0,edges[edge].to=u,edges[edge].next=head[v],head[v]=edge++;
}
bool Dinic_bfs(){
int i,u,v,l,r=0;
for (i=0; i<node; i++) dis[i]=-1;
dis[q[r++]=src]=0;
for (l=0; l<r; l++){
for (i=head[u=q[l]]; i!=-1; i=edges[i].next){
if (edges[i].flow&&dis[v=edges[i].to]<0){
dis[q[r++]=v]=dis[u]+1;
if (v==dest) return true;
}
}
}
return false;
}
LL Dinic_dfs(int u,LL exp){
if (u==dest) return exp;
int v;
LL tmp;
for (int &i=work[u]; i!=-1; i=edges[i].next){
if (edges[i].flow&&dis[v=edges[i].to]==dis[u]+1&&
(tmp=Dinic_dfs(v,min(exp,edges[i].flow)))>0){
edges[i].flow-=tmp;
edges[i^1].flow+=tmp;
return tmp;
}
}
return 0;
}
LL Dinic_flow(){
int i;
LL ret=0,delta;
while (Dinic_bfs()){
for (i=0;i<node;i++) work[i]=head[i];
while (delta=Dinic_dfs(src,INFF)) ret+=delta;
}
return ret;
}
void dfs(int u){
vis[u]=true;
for (int i=head[u];i!=-1;i=edges[i].next){
int v=edges[i].to;
if (vis[v]) continue;
if (edges[i].flow>0){
dfs(v);
cnt++;
}
}
}
}solver;
int n,m;
LL b[maxn];
LL sum;
int main()
{
while (~scanf("%d%d",&n,&m))
{
clr(vis,0);
sum=0;
solver.prepare(n+2,0,n+1);
REP_1(i,n){
scanf("%I64d",&b[i]);
if (b[i]>0) sum+=b[i];
if (b[i]>0) solver.addedge(0,i,b[i]);
if (b[i]<0) solver.addedge(i,n+1,-b[i]);
}
REP(i,m){
int x,y;
scanf("%d%d",&x,&y);
solver.addedge(x,y,INFF);
}
LL ans=sum-solver.Dinic_flow();
cnt=0;
solver.dfs(0);
printf("%d ",cnt);
printf("%I64d
",ans);
}
return 0;
}