poj 2987 Firing

题意：有n个员工，现在公司要裁员，告诉你裁掉每个员工可以得的到的盈利c，c大于0也可以小于0，现在问你公司的最大盈利为多少，并且在最大盈利的基础上至少炒掉多少人。最大闭合权图：建立超级源点和使盈利为正数的员工相连，权值为该员工给公司带来盈利，建立超级汇点和使公司盈利为负的员工相连,权值为该员工给公司带来盈利的绝对值，存在从属关系的员工之间连一条边，全值为inf，跑一遍最大流算法即可。#include<stdio.h>   #include<string.h>   #include<iostream>   #define N 200000   #define inf 1000000000   using namespace std;  struct Edge{  long long u,v,c,next;  }edge[1000010];  long long head[80010],pre[80010],cur[1000010],dis[80010],gap[1000010],vis[80010];  long long start,end,n,m,e,a,b,c,sum,num;    void add_adge(long long u,long long v,long long c1)  {     edge[e].u=u; edge[e].v=v; edge[e].c=c1; edge[e].next=head[u]; head[u]=e++;     edge[e].u=v; edge[e].v=u; edge[e].c=0;  edge[e].next=head[v]; head[v]=e++;  }  void dfs(long long u) {     vis[u]=1;     for(long long i=head[u]; i!=-1;i=edge[i].next)      {                    if(edge[i].c>0&&!vis[edge[i].v])          {             num++;             dfs(edge[i].v);         }     }}long long sap()    {         long long  flow=0,aug=inf,u;        long long flag;       for(long long i=0; i<=n+1; i++)        {            cur[i]=head[i];            gap[i]=dis[i]=0;        }        gap[start]=n+2;         u=pre[start]=start;       while(dis[start]<n+2)        {            flag=0;            for(long long &j=cur[u]; j!=-1; j=edge[j].next)            {                long long v=edge[j].v;                if(edge[j].c>0&&dis[u]==dis[v]+1)                {                    flag=1;                    if(edge[j].c<aug) aug=edge[j].c;                    pre[v]=u;                    u=v;                    if(u==n+1)                    {                        flow+=aug;                        while(u!=0)                        {                            u=pre[u];                            edge[cur[u]].c-=aug;                          edge[cur[u]^1].c+=aug;                     }                        aug=inf;                    }                    break;                }            }           if(flag) continue;            long long mindis=n+2;            for(long long j=head[u]; j!=-1; j=edge[j].next)            {                long long v=edge[j].v;                if(edge[j].c>0&&dis[v]<mindis)                {                    mindis=dis[v];                    cur[u]=j;                }            }            if((--gap[dis[u]])==0)                break;            gap[dis[u]=mindis+1]++;            u=pre[u];        }        return flow;    }    int main()  {     while(cin>>n>>m)   {          e=0;start=0;num=0;end=n+1;         long long sum=0;        memset(head,-1,sizeof(head));          memset(vis,0,sizeof(vis));        for(long long  i=1;i<=n;i++)          {             cin>>c;           if(c>0)            {              add_adge(0,i,c);              sum += c;           }else if(c<0) add_adge(i,n+1,-c);         }          for(long long i=1;i<=m;i++)          {             cin>>a>>b;            add_adge(a,b,inf);           }         long long ans=sap();        sum -= ans;        dfs(0);        //printf("%lld\n",sum);        printf("%lld %I64d\n",num,sum);   }     return 0;  }