题目连接: (luogu) https://www.luogu.org/problemnew/show/P2604

(bzoj) https://www.lydsy.com/JudgeOnline/problem.php?id=1834

题解: 第一问所有的费用全按\(0\)建，跑完了之后很自然想到利用残余网络。

把\(n\)和一个新点\(T\)连边，然后原来的残量网络保留，在此基础上对于原来的每条边流量均按\(+\inf\)建，费用为原始费用再跑一遍即可。

时间复杂度\(O(MaxFlowMinCost(n,m))\)

（然而智障的我不会处理残量网路还想做\(K\)次费用流每次\(+1\)，真是没脑子）

代码

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define llong long long
using namespace std;const int N = 1002;
const int M = 10000;
const llong INF = 1000000000000ll;
struct Edge
{int u,v,nxt,rev; llong c,w;
} e[(M<<1)+3];
Edge ae[M+3];
int fe[N+3];
int que[N+3];
llong dis[N+3];
bool inq[N+3];
int lst[N+3];
int n,m,en,p,s,t;
llong mf,mc;void addedge(int u,int v,llong w,llong c)
{en++; e[en].u = u; e[en].v = v; e[en].w = w; e[en].c = c;e[en].nxt = fe[u]; fe[u] = en; e[en].rev = en+1;en++; e[en].u = v; e[en].v = u; e[en].w = 0; e[en].c = -c;e[en].nxt = fe[v]; fe[v] = en; e[en].rev = en-1;
}bool spfa()
{for(int i=1; i<=n; i++) dis[i] = INF;int head = 1,tail = 2; que[tail-1] = s; dis[s] = 0ll;while(head!=tail){int u = que[head]; head++; if(head==n+1) head = 1;for(int i=fe[u]; i; i=e[i].nxt){if(e[i].w>0 && dis[e[i].v]>dis[u]+e[i].c){dis[e[i].v] = dis[u]+e[i].c;lst[e[i].v] = i;if(!inq[e[i].v]){inq[e[i].v] = true;que[tail] = e[i].v; tail++; if(tail==n+1) tail = 1;}}}inq[u] = false;}return dis[t]!=INF;
}void calcflow()
{llong flow = INF;for(int i=t; i!=s; i=e[lst[i]].u){flow = min(flow,e[lst[i]].w);}for(int i=t; i!=s; i=e[lst[i]].u){e[lst[i]].w -= flow; e[e[lst[i]].rev].w += flow;}mf += flow; mc += flow*dis[t];
}void mfmc()
{while(spfa()){calcflow();}
}int main()
{scanf("%d%d%d",&n,&m,&p);for(int i=1; i<=m; i++){scanf("%d%d%lld%lld",&ae[i].u,&ae[i].v,&ae[i].w,&ae[i].c);addedge(ae[i].u,ae[i].v,ae[i].w,0ll);}s = 1; t = n; mf = mc = 0ll;mfmc();printf("%lld ",mf);n++; addedge(n-1,n,p,0); t = n;for(int i=1; i<=m; i++){addedge(ae[i].u,ae[i].v,INF,ae[i].c);}mf = mc = 0ll;mfmc();printf("%lld\n",mc);return 0;
}