POJ 2987 Firing
2018-06-27 10:02:38来源:未知 阅读 ()
Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions: 10804 | Accepted: 3263 |
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
Source
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<queue> 6 #define lli long long int 7 using namespace std; 8 const int MAXN=2000001; 9 const int INF = 1e8; 10 inline void read(int &n) 11 { 12 char c='+';int x=0;bool flag=0; 13 while(c<'0'||c>'9'){c=getchar();if(c=='-')flag=1;} 14 while(c>='0'&&c<='9'){x=x*10+c-48;c=getchar();} 15 n=flag==1?-x:x; 16 } 17 int n,m,s,t; 18 struct node 19 { 20 int u,v,flow,nxt; 21 }edge[MAXN]; 22 int head[MAXN]; 23 int cur[MAXN]; 24 int num=0; 25 int deep[MAXN]; 26 lli tot=0; 27 bool vis[MAXN]; 28 lli out; 29 void add_edge(int x,int y,int z) 30 { 31 edge[num].u=x; 32 edge[num].v=y; 33 edge[num].flow=z; 34 edge[num].nxt=head[x]; 35 head[x]=num++; 36 } 37 void add(int x,int y,int z) 38 { 39 add_edge(x,y,z); 40 add_edge(y,x,0); 41 } 42 bool BFS() 43 { 44 memset(deep,0,sizeof(deep)); 45 deep[s]=1; 46 queue<int>q; 47 q.push(s); 48 while(q.size()!=0) 49 { 50 int p=q.front(); 51 q.pop(); 52 for(int i=head[p];i!=-1;i=edge[i].nxt) 53 if(!deep[edge[i].v]&&edge[i].flow) 54 deep[edge[i].v]=deep[edge[i].u]+1, 55 q.push(edge[i].v); 56 } 57 return deep[t]; 58 59 } 60 lli DFS(int now,int nowflow) 61 { 62 if(now==t||nowflow<=0) 63 return nowflow; 64 lli totflow=0; 65 for(int &i=cur[now];i!=-1;i=edge[i].nxt) 66 { 67 if(deep[edge[i].v]==deep[edge[i].u]+1&&edge[i].flow) 68 { 69 int canflow=DFS(edge[i].v,min(nowflow,edge[i].flow)); 70 edge[i].flow-=canflow; 71 edge[i^1].flow+=canflow; 72 totflow+=canflow; 73 nowflow-=canflow; 74 if(nowflow<=0) 75 break; 76 } 77 78 } 79 return totflow; 80 } 81 void Dinic() 82 { 83 lli ans=0; 84 while(BFS()) 85 { 86 memcpy(cur,head,MAXN); 87 ans+=DFS(s,1e8); 88 } 89 out=tot-ans; 90 } 91 int find_pep(int now) 92 { 93 vis[now]=1; 94 for(int i=head[now];i!=-1;i=edge[i].nxt) 95 if(!vis[edge[i].v]&&edge[i].flow) 96 find_pep(edge[i].v); 97 } 98 int main() 99 { 100 //freopen("a.in","r",stdin); 101 //freopen("c.out","w",stdout); 102 while(~scanf("%d%d",&n,&m)) 103 { 104 memset(head,-1,sizeof(head)); 105 num=0; 106 s=0,t=n+1; 107 tot=0; 108 for(int i=1;i<=n;i++) 109 { 110 int a;read(a); 111 if(a>0) tot+=a,add(s,i,a); 112 if(a<0) add(i,t,-a); 113 } 114 for(int i=1;i<=m;i++) 115 { 116 int x,y;read(x);read(y); 117 add(x,y,INF); 118 } 119 Dinic(); 120 memset(vis,0,sizeof(vis)); 121 int ans=0; 122 find_pep(s); 123 for(int i=1;i<=n;i++) 124 ans+=vis[i]; 125 //cout<<ans<<" "<<out<<endl; 126 printf("%d %I64d\n",ans,out); 127 } 128 return 0; 129 }
标签:
版权申明:本站文章部分自网络,如有侵权,请联系:west999com@outlook.com
特别注意:本站所有转载文章言论不代表本站观点,本站所提供的摄影照片,插画,设计作品,如需使用,请与原作者联系,版权归原作者所有
- POJ-3278 2020-04-01
- Asteroids!_poj2225 2020-02-09
- poj-1753题题解思路 2020-01-26
- POJ1852 2019-11-11
- POJ2431 优先队列+贪心 - biaobiao88 2019-11-03
IDC资讯: 主机资讯 注册资讯 托管资讯 vps资讯 网站建设
网站运营: 建站经验 策划盈利 搜索优化 网站推广 免费资源
网络编程: Asp.Net编程 Asp编程 Php编程 Xml编程 Access Mssql Mysql 其它
服务器技术: Web服务器 Ftp服务器 Mail服务器 Dns服务器 安全防护
软件技巧: 其它软件 Word Excel Powerpoint Ghost Vista QQ空间 QQ FlashGet 迅雷
网页制作: FrontPages Dreamweaver Javascript css photoshop fireworks Flash