洛谷P4016 负载平衡问题 费用流

#include<cstdio>              //好题
#include<algorithm>
#include<vector>
#include<cstring>
#include<queue>
using namespace std;
const int maxn=500;
const int INF=1000000+23666;
typedef long long ll;
int A[maxn];
int s,t,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 addedge(int u,int v,int c,int f){
    	edges.push_back(Edge(u,v,c,f));    //正向弧
    	edges.push_back(Edge(v,u,0,-f));   //反向弧
    	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 maxflow(){
        while(SPFA());
        return ans;
    }
   // ll getcost(){return ans;}
}op;
int main()
{
     int N;scanf("%d",&N);
     s=0,t=N+1,n=N+1;
     int sum=0,ave;
     for(int i=1;i<=N;++i){
          int a;scanf("%d",&a);
          A[i]=a;
          sum+=a;
     }
     ave=sum/N;
     for(int i=1;i<=N;++i)
     {
          int a=A[i]-ave;
          if(a>0)op.addedge(s,i,a,0);
          if(a<0)op.addedge(i,t,-a,0);
     }
     for(int i=2;i<N;++i)
     {
     	     op.addedge(i,i-1,INF,1);
     	     op.addedge(i,i+1,INF,1);
     }
     op.addedge(1,2,INF,1);
     op.addedge(1,N,INF,1);
     op.addedge(N,N-1,INF,1);
     op.addedge(N,1,INF,1);
     printf("%lld",op.maxflow());
     return 0;
}