学步园

费用流

混合欧拉回路

#include<cstdio>
#include<cstring>
#include<queue>
#define N 300
#define inf 100000000
using namespace std;
typedef long long ll;
int n,cnt;
int head[N],pre[N],vis[N],in[N],out[N];
int dis[N];
int ans,sum,suma,sumb;
struct Edge{
    int u,v,c,next;
    ll w;
}edge[8010];

void init(){
    memset(head,-1,sizeof(head));
    memset(in,0,sizeof(in));
    memset(out,0,sizeof(out));
    cnt=sum=ans=suma=sumb=0;
}
void addedge(int u,int v,int cap,int cost){
    edge[cnt].u=u;
    edge[cnt].v=v;
    edge[cnt].c=cap;
    edge[cnt].w=cost;
    edge[cnt].next=head[u];
    head[u]=cnt++;
    edge[cnt].u=v;
    edge[cnt].v=u;
    edge[cnt].c=0;
    edge[cnt].w=-cost;
    edge[cnt].next=head[v];
    head[v]=cnt++;
}
bool spfa(){
    int i,j;
    queue<int>q;
    q.push(0);
    for(i=1;i<=n;i++){
        dis[i]=inf;
        vis[i]=0;
    }
    dis[0]=0,vis[0]=1;
    while(!q.empty()){
        int u=q.front();
        q.pop();
        vis[u]=0;
        for(i=head[u];i!=-1;i=edge[i].next){
            int v=edge[i].v;
            if(edge[i].c && dis[v] > dis[u]+edge[i].w){
                dis[v]=dis[u]+edge[i].w;
                pre[v]=i;
                if(!vis[v]){
                    vis[v]=1;
                    q.push(v);
                }
            }
        }
    }
    if(dis[n]==inf)return 0;
    return 1;
}
void end(){
    int u,p,cap=-1;
    for(u=n;u;u=edge[p].u){
        p=pre[u];
        if(cap==-1 || cap>edge[p].c)
            cap=edge[p].c;
    }
    suma+=cap;
    for(u=n;u;u=edge[p].u){
        p=pre[u];
        edge[p].c-=cap;
        edge[p^1].c+=cap;
        ans+=cap*edge[p].w;
    }
}
int main(){
    int t,T,i;
    int v,e,s,tt;
    int vv,u,a,b;
    scanf("%d",&T);
    for(t=1;t<=T;t++){
        scanf("%d %d %d %d",&v,&e,&s,&tt);
        init();
        for(i=1;i<=e;i++){
            scanf("%d %d %d %d",&u,&vv,&a,&b);
            if(a<=b){
                addedge(u,vv,1,b-a);   //建这条边,如果费用流经过这条边,那么u的out++,in-- 并且vv的in++,out--
                in[vv]++,out[u]++;
                sum+=a;
            }
            else{
                addedge(vv,u,1,a-b);   //建这条边,如果费用流经过这条边,那么vv的out++，u的in++
                sum+=b;
            }
        }
        n=v+1;
        in[s]++,out[tt]++;
        for(i=1;i<=v;i++)
            if(in[i]<out[i]){
                addedge(0,i,out[i]-in[i],0);
                sumb+=out[i]-in[i];
            }
            else if(in[i]>out[i])
                addedge(i,n,in[i]-out[i],0);
        while(spfa())
            end();
        if(suma==sumb)
            printf("Case %d: %d\n",t,ans+sum);
        else
            printf("Case %d: impossible\n",t);
    }
    return 0;
}