loj2100 「TJOI2015」线性代数

先推公式，推出个这，然后因为是 \(0/1\) 矩阵，选一个有损耗，两个一组有加成，就想到了最大权闭合子图，（飞行计划问题）

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;
int n, uu, ss, tt, hea[250505], cnt, cur[250505], maxFlow, lev[250505], ans;
const int oo=0x3f3f3f3f;
queue<int> d;
struct Edge{
	int too, nxt, val;
}edge[1600005];
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));
	lev[ss] = 1;
	d.push(ss);
	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=cur[x]; i!=-1; 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;
			if(addFlow==lim)	break;
		}
	}
	return addFlow;
}
void dinic(){
	while(bfs()){
		for(int i=ss; i<=tt; i++)	cur[i] = hea[i];
		maxFlow += dfs(ss, oo);
	}
}
int main(){
	memset(hea, -1, sizeof(hea));
	cin>>n;
	ss = 0; tt = n + n * n + 1;
	for(int i=1; i<=n; i++)
		for(int j=1; j<=n; j++){
			scanf("%d", &uu);
			ans += uu;
			int p=(i-1)*n+j;
			addEdge(ss, p, uu);
			addEdge(p, n*n+i, oo);
			addEdge(p, n*n+j, oo);
		}
	for(int i=1; i<=n; i++){
		scanf("%d", &uu);
		addEdge(n*n+i, tt, uu);
	}
	dinic();
	cout<<ans-maxFlow<<endl;
	return 0;
}