HOJ 2634 网络流 【蕴含式最大获利问题】

参考：http://wenku.baidu.com/link?url=ZO8hOpUlTb04rmoob64R-OohRZfyw-xv7cMdwKX-_RXsUKnExnJSxlYuEnnRqF3bR8hIYJuKjrgiN38cIBXjtNi_7y-FRy7rZRc_vGP9Bj_

【蕴含式最大获利问题】

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
const int MAXN=210,inf=0x3f3f3f3f;
struct ISAP
{   struct Edge
{
int from,to,cap,flow;
Edge(){}
Edge(int a,int b,int c,int d):from(a),to(b),cap(c),flow(d){}
};
int n,m,s,t;//结点数，边数(含反向弧)，源点，汇点
vector<Edge> edges;//边表，edges[e]&edges[e^1]互为反向弧
vector<int> G[MAXN];//邻接表，G[i][j]表示结点i的第j条边在e数组中的序号
bool vis[MAXN];//BFS使用
int d[MAXN];//从起点到i的距离
int cur[MAXN];//当前弧下标
int p[MAXN];//可增广路上的上一条弧
int num[MAXN];//距离标号计数

void AddEdge(int from,int to,int cap)//重边不影响
{
edges.push_back(Edge(from,to,cap,0));
edges.push_back(Edge(to,from,0,0));//容量为0,表示反向弧
m=edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}

void init(int n)
{
this->n=n;
for(int i=0;i<n;++i) G[i].clear();
edges.clear();
}

void BFS()//反向
{
memset(vis,0,sizeof(vis));
queue<int> Q;
Q.push(t);
d[t]=0;
vis[t]=1;
while(!Q.empty())
{
int x=Q.front();
Q.pop();
for(int i=0; i<G[x].size(); ++i)
{
Edge& e=edges[G[x][i]^1];
if(!vis[e.from]&&e.cap>e.flow)
{
vis[e.from]=1;
d[e.from]=d[x]+1;
Q.push(e.from);
}
}
}
}

int Augment()
{
int x=t,a=inf;
while(x!=s)
{
Edge& e=edges[p[x]];
a=min(a,e.cap-e.flow);
x=edges[p[x]].from;
}
x=t;
while(x!=s)
{
edges[p[x]].flow+=a;
edges[p[x]^1].flow-=a;
x=edges[p[x]].from;
}
return a;
}

int Maxflow(int s,int t)//结点数
{
this->s=s,this->t=t;
int flow=0;
BFS();
memset(num,0,sizeof(num));
for(int i=0;i<n;++i) ++num[d[i]];
int x=s;
memset(cur,0,sizeof(cur));
while(d[s]<n)
{
if(x==t)
{
flow+=Augment();
x=s;
}
int ok=0;
for(int i=cur[x];i<G[x].size();++i)
{
Edge& e=edges[G[x][i]];
if(e.cap>e.flow&&d[x]==d[e.to]+1)//Advance
{
ok=1;
p[e.to]=G[x][i];
cur[x]=i;
x=e.to;
break;
}
}
if(!ok)//Retreat
{
int m=n-1;
for(int i=0;i<G[x].size();++i)
{
Edge& e=edges[G[x][i]];
if(e.cap>e.flow) m=min(m,d[e.to]);
}
if(--num[d[x]]==0) break;//gap优化
num[d[x]=m+1]++;
cur[x]=0;
if(x!=s) x=edges[p[x]].from;
}
}
return flow;
}
void dfs1(int x)
{   vis[x]=1;
for(int i=0;i<G[x].size();++i)
{
Edge& e=edges[G[x][i]];
if(e.flow<e.cap&&!vis[e.to]) dfs1(e.to);
}
}
void dfs2(int x)
{   vis[x]=1;
for(int i=0;i<g[x].size();++i)
if(!vis[g[x][i]]) dfs2(g[x][i]);
}
vector<int> g[MAXN];
void solve(int y)
{   for (int i=0;i<n;i++) g[i].clear();
for (int i=0;i<edges.size();i++)
if (edges[i].flow<edges[i].cap)
g[edges[i].to].push_back(edges[i].from);
memset(vis,0,sizeof(vis));
dfs1(s);
//dfs2(t);
printf("project:\n");
for (int i=1;i<=y;i++)
if (vis[i]) printf("%d ",i);
printf("\nemployer:\n");
for (int i=y+1;i<n-1;i++)
if (vis[i]) printf("%d ",i-y);
}
}it;
int n,m;
void doit()
{   scanf("%d%d",&n,&m);
int t=n+m+1,tot=0;
it.init(t+1);
for (int i=1,x;i<=n;i++)
{
scanf("%d",&x); tot+=x;
it.AddEdge(0,i,x);
}
for (int i=1,x;i<=m;i++)
{
scanf("%d",&x);
it.AddEdge(n+i,t,x);
}
for (int i=1,x,y;i<=n;i++)
{
scanf("%d",&x);
for (int j=1;j<=x;j++)
{
scanf("%d",&y); y++;
it.AddEdge(i,n+y,inf);
}
}
printf("%d\n",tot-it.Maxflow(0,t));
//it.solve(n);
}

int main()
{   int cas;
scanf("%d",&cas);
while (cas--) doit();
return 0;
}