第一问直接跑最大流即可。建图的时候按照费用流建，费用为0.
对于第二问，在第一问dinic剩下的残量网络上建图，对原图的每条边(i,j)，建(i,j,inf,cij)，表示可以用c的花费增广这条路。然后从新建一个源点，连(s,1,k,0)表示要增加k的流量。跑最小费用最大流即可。
#include<iostream>#include<cstdio>#include<cstring>#include<queue>using namespace std;const int N=1000005,inf=1e9;int n,m,k,h[N],cnt=1,le[N],s,t,ans,dis[N],fr[N],c[N];bool v[N];struct qwe{ ???int ne,no,to,va,c;}e[N];int read(){ ???int r=0,f=1; ???char p=getchar(); ???while(p>‘9‘||p<‘0‘) ???{ ???????if(p==‘-‘) ???????????f=-1; ???????p=getchar(); ???} ???while(p>=‘0‘&&p<=‘9‘) ???{ ???????r=r*10+p-48; ???????p=getchar(); ???} ???return r*f;}void add(int u,int v,int w,int c){ ???cnt++; ???e[cnt].ne=h[u]; ???e[cnt].no=u; ???e[cnt].to=v; ???e[cnt].va=w; ???e[cnt].c=c; ???h[u]=cnt;}void ins(int u,int v,int w,int c){ ???add(u,v,w,c); ???add(v,u,0,-c);}bool bfs(){ ???queue<int>q; ???memset(le,0,sizeof(le)); ???le[s]=1; ???q.push(s); ???while(!q.empty()) ???{ ???????int u=q.front(); ???????q.pop(); ???????for(int i=h[u];i;i=e[i].ne) ???????????if(e[i].va>0&&!le[e[i].to]) ???????????{ ???????????????le[e[i].to]=le[u]+1; ???????????????q.push(e[i].to); ???????????} ???} ???return le[t];}int dfs(int u,int f){ ???if(!f||u==t) ???????return f; ???int us=0; ???for(int i=h[u];i&&us<f;i=e[i].ne) ???????if(e[i].va>0&&le[e[i].to]==le[u]+1) ???????{ ???????????int t=dfs(e[i].to,min(e[i].va,f-us)); ???????????e[i].va-=t; ???????????e[i^1].va+=t; ???????????us+=t; ???????} ???if(!us) ???????le[u]=0; ???return us;}int dinic(){ ???int re=0; ???while(bfs()) ???????re+=dfs(s,inf); ???return re;}bool spfa(){ ???queue<int>q; ???for(int i=s;i<=t;i++) ???????dis[i]=inf; ???dis[s]=0; ???v[s]=0; ???q.push(s); ???while(!q.empty()) ???{ ???????int u=q.front(); ???????q.pop(); ???????v[u]=0; ???????for(int i=h[u];i;i=e[i].ne) ???????????if(e[i].va>0&&dis[e[i].to]>dis[u]+e[i].c) ???????????{ ???????????????dis[e[i].to]=dis[u]+e[i].c; ???????????????fr[e[i].to]=i; ???????????????if(!v[e[i].to]) ???????????????{ ???????????????????v[e[i].to]=1; ???????????????????q.push(e[i].to); ???????????????} ???????????} ???} ???return dis[t]!=inf;}void mcf(){ ???int x=inf; ???for(int i=fr[t];i;i=fr[e[i].no]) ???????x=min(x,e[i].va); ???for(int i=fr[t];i;i=fr[e[i].no]) ???{ ???????e[i].va-=x; ???????e[i^1].va+=x; ???????ans+=e[i].c*x; ???}}int main(){ ???n=read(),m=read(),k=read(); ???s=1,t=n; ???for(int i=1;i<=m;i++) ???{ ???????int x=read(),y=read(),z=read();c[i]=read(); ???????ins(x,y,z,0); ???} ???printf("%d ",dinic()); ???s=0,t=n; ???int now=cnt; ???for(int i=2;i<=now;i+=2) ???????ins(e[i].no,e[i].to,inf,c[i>>1]); ???ins(s,1,k,0); ???while(spfa()) ???????mcf(); ???printf("%d\n",ans); ???return 0;}