luogu2762 太空飞行计划问题

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("\n");
	for(int i=n+1; i<=n+m; i++)
		if(lev[i])	printf("%d ", i-n);
	printf("\n");
	cout<<tot<<endl;
	return 0;
}