Acme Corporation UVA - 11613 费用流

Posted 2021-02-09 guangheli

Code:

#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
const int maxn=500;
const int INF=10000000+23666;
typedef long long ll;
int s,t,n;
struct Edge{
	int from,to,cost;
	ll cap;
	Edge(int u,int v,ll c,int f):from(u),to(v),cap(c),cost(f){}
};
struct MCMF{
    vector<Edge>edges;
    vector<int>G[maxn];
    ll d[maxn],inq[maxn],a[maxn],flow2[maxn];
    queue<int>Q;
    ll ans=0;
    int flow=0;
    void init(){
    	   memset(a,0,sizeof(a));
    	   edges.clear();
    	   for(int i=0;i<maxn;++i)G[i].clear();
    	   ans=0,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]>=0)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;
    }
}op;
int main()
{
	int T;scanf("%d",&T);
	for(int cas=1;cas<=T;++cas)
	{
          op.init();int cnt=0;
		int m,I;scanf("%d%d",&m,&I);
		s=0,n=2*m+1,t=n;
		for(int i=1;i<=m;++i){
			cnt+=2;
		     int mi,ni,pi,si,E;
		     scanf("%d%d%d%d%d",&mi,&ni,&pi,&si,&E);
               op.addedge(s,cnt,ni,mi);
               op.addedge(cnt-1,t,si,-pi);
               int a=cnt-1;
               for(int j=0;j<=E&&a<t;++j){
                    op.addedge(cnt,a,INF,j*I);
                    a+=2;
               }
          }
          printf("Case %d: %lld
",cas,op.maxflow());  
	}
	return 0;
}