洛谷P4014 分配问题 费用流

Code：

#include<vector>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int maxn=1003;
const int INF=1000000+23666;
typedef long long ll;
int A[200][200];
int s,t,n;
int N;
struct Edge{
	int from,to,cap,cost;
	Edge(int u,int v,int c,int f):from(u),to(v),cap(c),cost(f){}
};
struct MCMF{
	vector<Edge>edges;
	vector<int>G[maxn];
	int d[maxn],inq[maxn],a[maxn],flow2[maxn];
	queue<int>Q;
	ll ans=0;
	int flow=0;
	void init(){
		ans=0,flow=0;
		for(int i=0;i<maxn;++i)G[i].clear();
		for(int i=0;i<maxn;++i)edges.clear();
	}
	void addedge(int u,int v,int c,int f){
		edges.push_back(Edge(u,v,c,f));    
		edges.push_back(Edge(v,u,0,-f));   
		int m=edges.size();
		G[u].push_back(m-2);
		G[v].push_back(m-1);
	}
	int SPFA(){
		for(int i=0;i<=n;++i)d[i]=INF,flow2[i]=INF;
		memset(inq,0,sizeof(inq));int f=INF;
		d[s]=0,inq[s]=1;Q.push(s);
		while(!Q.empty()){
			int u=Q.front();Q.pop();inq[u]=0;
			int sz=G[u].size();
			for(int i=0;i<sz;++i){
				  Edge e=edges[G[u][i]];
				  if(e.cap>0&&d[e.to]>d[u]+e.cost){
					  a[e.to]=G[u][i];
					  d[e.to]=d[u]+e.cost;
					  flow2[e.to]=min(flow2[u],e.cap);
					  if(!inq[e.to]){inq[e.to]=1;Q.push(e.to);}
				  }
			}
		}
		if(d[t]==INF)return 0;
		f=flow2[t];
		flow+=f;
		int u=edges[a[t]].from;
		edges[a[t]].cap-=f;
		edges[a[t]^1].cap+=f;
		while(u!=s){
			edges[a[u]].cap-=f;
			edges[a[u]^1].cap+=f;
			u=edges[a[u]].from;
		}
		ans+=(ll)(d[t]*f);
		return 1;
	}
	ll  getcost(){
		while(SPFA());
		return ans;
	}
}op;
void build(int mul){
	op.init();
	for(int i=1;i<=N;++i)
		for(int j=1;j<=N;++j)
		{
			int a=j,b=i+N;
			op.addedge(a,b,1,A[i][j]*mul);
		}
	for(int i=1;i<=N;++i)op.addedge(s,i,1,0);
	for(int i=N+1;i<=N+N;++i)op.addedge(i,t,1,0);
}
int main(){
	scanf("%d",&N);
	for(int i=1;i<=N;++i)
		for(int j=1;j<=N;++j)scanf("%d",&A[i][j]);
	        s=0,t=N+N+1,n=N+N+1;
            build(1);
            printf("%lld\n",op.getcost());
            build(-1);
            printf("%lld",op.getcost()*(-1));
            return 0;
}