BZOJ 1834 Luogu P2604 [ZJOI2010]网络扩容 (最小费用最大流)

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;
}