最大权闭合子图
参考这，胡伯涛论文。
10，8，6，3这个简单割对应的闭合子图是A1，B1，B2
输出路径时，最后一次层次图中，与源点相连的点即选做的实验，与汇点相连的点即选用的仪器。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;
int n, m, maxFlow, uu, vv, ss, tt, tot, cnt, hea[105], lev[105], fff;
const int oo=0x3f3f3f3f;
queue<int> d;
struct Edge{
int too, nxt, val;
}edge[20005];
char aaa[100005];
void add_edge(int fro, int too, int val){
edge[cnt].nxt = hea[fro];
edge[cnt].too = too;
edge[cnt].val = val;
hea[fro] = cnt++;
}
void addEdge(int fro, int too, int val){
add_edge(fro, too, val);
add_edge(too, fro, 0);
}
bool bfs(){
memset(lev, 0, sizeof(lev));
d.push(ss);
lev[ss] = 1;
while(!d.empty()){
int x=d.front();
d.pop();
for(int i=hea[x]; i!=-1; i=edge[i].nxt){
int t=edge[i].too;
if(!lev[t] && edge[i].val>0){
lev[t] = lev[x] + 1;
d.push(t);
}
}
}
return lev[tt]!=0;
}
int dfs(int x, int lim){
if(x==tt) return lim;
int addFlow=0;
for(int i=hea[x]; i!=-1 && addFlow<lim; i=edge[i].nxt){
int t=edge[i].too;
if(lev[t]==lev[x]+1 && edge[i].val>0){
int tmp=dfs(t, min(lim-addFlow, edge[i].val));
edge[i].val -= tmp;
edge[i^1].val += tmp;
addFlow += tmp;
}
}
return addFlow;
}
void dinic(){
while(bfs()) maxFlow += dfs(ss, oo);
}
int main(){
memset(hea, -1, sizeof(hea));
cin>>n>>m;
ss = 0; tt = n + m + 1;
for(int i=1; i<=n; i++){
scanf("%d", &uu);
tot += uu;
addEdge(ss, i, uu);
fff = 0;
char ch=getchar();
memset(aaa, 0, sizeof(aaa));
while((ch>='0' && ch<='9') || ch==' ')
aaa[++fff] = ch, ch = getchar();
for(int j=1; j<=fff+1; j++){
if(aaa[j]>='0' && aaa[j]<='9')
vv = vv * 10 + aaa[j] - '0';
else{
vv += n;
if(vv>n) addEdge(i, vv, oo);
vv = 0;
}
}
}
for(int i=1; i<=m; i++){
scanf("%d", &uu);
addEdge(n+i, tt, uu);
}
dinic();
tot -= maxFlow;
for(int i=1; i<=n; i++)
if(lev[i]) printf("%d ", i);
printf("
");
for(int i=n+1; i<=n+m; i++)
if(lev[i]) printf("%d ", i-n);
printf("
");
cout<<tot<<endl;
return 0;
}