思路

简单的费用流问题，跑出第一问后在残量网络上加边求最小费用即可

代码

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
const int MAXN = 4000;
const int MAXM = 10000;
const int INF = 0x3f3f3f3f;
struct Edge{int u,v,cap,cost,flow;
};
struct data{int u,v,c,w;
}E[MAXM];
int s,t,n,m,k;
vector<Edge> edges;
vector<int> G[MAXN];
void addedge(int u,int v,int cap,int cost){edges.push_back((Edge){u,v,cap,cost,0});edges.push_back((Edge){v,u,0,-cost,0});int cnt=edges.size();G[u].push_back(cnt-2);G[v].push_back(cnt-1);
}
int a[MAXN],d[MAXN],p[MAXN],vis[MAXN];
queue<int> q;
bool SPFA(int &flow,int &cost){memset(d,0x3f,sizeof(d));memset(p,0,sizeof(p));q.push(s);vis[s]=true;a[s]=INF;d[s]=0;while(!q.empty()){int x=q.front();q.pop();vis[x]=false;for(int i=0;i<G[x].size();i++){Edge &e = edges[G[x][i]];if(e.cap>e.flow&&d[x]+e.cost<d[e.v]){d[e.v]=d[x]+e.cost;a[e.v]=min(a[x],e.cap-e.flow);p[e.v]=G[x][i];if(!vis[e.v]){vis[e.v]=true;q.push(e.v);}}}}if(d[t]==INF)return false;cost+=d[t]*a[t];flow+=a[t];for(int i=t;i!=s;i=edges[p[i]].u){edges[p[i]].flow+=a[t];edges[p[i]^1].flow-=a[t];}return true;
}
void MCMF(int &flow,int &cost){flow=0,cost=0;while(SPFA(flow,cost));
}
int main(){scanf("%d %d %d",&n,&m,&k);s=1,t=n;for(int i=1;i<=m;i++){scanf("%d %d %d %d",&E[i].u,&E[i].v,&E[i].c,&E[i].w);addedge(E[i].u,E[i].v,E[i].c,0);}int flow=0,cost=0;MCMF(flow,cost);printf("%d ",flow);s=MAXN-2;addedge(s,1,k,0);for(int i=1;i<=m;i++){addedge(E[i].u,E[i].v,INF,E[i].w);}MCMF(flow,cost);printf("%d\n",cost);return 0;
}