BZOJ 1834 ZJOI2010 network network expansion

1834: [ZJOI2010]Network network expansion

Time Limit: 3 Sec  Memory Limit: 64 MB
Submit: 3735  Solved: 2001
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Description

Given a directed graph, each edge has a capacity C and an expansion cost W. The expansion cost here refers to the cost required to expand the capacity by 1.

beg: 

1. In the case of no expansion, the maximum flow from 1 to N; 

2. The minimum expansion cost required to increase the maximum flow from 1 to N by K.

Input

The first line contains three integers N, M, K, which represent the number of points, edges, and the amount of traffic that needs to be increased in the directed graph. 

The next M lines each contain four integers u, v, C, and W, representing an edge from u to v with a capacity of C and an expansion cost of W.

N<=1000，M<=5000，K<=10

Output

The output file contains two integers on one line, representing the answers to question 1 and question 2, respectively.

Sample Input

5 8 2
1 2 5 8
2 5 9 9
5 1 6 2
5 1 1 8
1 2 8 7
2 5 4 9
1 2 1 1
1 4 2 1

Sample Output

13 19

HINT

 

Source

 

Big water, the first question is the maximum flow, dinic run

The second question, we build a map on the residual network, connect the edges given by the question, the traffic inf, and the cost is the cost given by the question

The remaining side of the dinic run does not need to be moved, the cost is 0, and s to 1 is connected to an edge with a flow of k and a cost of 0. Just run the minimum cost and the maximum flow.

#include <bits/stdc++.h>
#define ll long long
#define inf 1e9+10
using namespace std;
inline int read(){
	int x=0;int f=1;char ch=getchar();
	while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
	while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
const int MAXN=1e5+10;
struct node{
	int y,next,back,flow,w;
}e[MAXN];
int len,linkk[MAXN],head,tail,level[1100],s,t,n,m,x[MAXN],y[MAXN],f[MAXN],c[MAXN],dis[1100],ans,vis[1100],q[MAXN],k;
inline void insert(int x,int y,int f,int c){
	e [++ len] .y = y; e [len] .next = linkk [x]; linkk [x] = len; e [len] .flow = f; e [len] .back = len + 1; e [len] .w = c;
	e [++ len] .y = x; e [len] .next = linkk [y]; linkk [y] = len; e [len] .flow = 0; e [len] .back = len-1; e [len] .w = -c;
}
namespace dinicc{
	inline bool getlevel(){
		head=tail=0;
		memset(level,-1,sizeof(level));
		level[s]=0;q[++tail]=s;
		while(head<tail){
			int tn=q[++head];
			for(int i=linkk[tn];i;i=e[i].next){
				if(level[e[i].y]==-1&&e[i].flow){
					level[e[i].y]=level[tn]+1;
					q[++tail]=e[i].y;
				}
			}
		}
		return level[t]>=0;
	}
	inline int getmaxflow(int x,int flow){
		if(x==t) return flow;
		int f=0,d;
		for(int i=linkk[x];i;i=e[i].next){
			if(e[i].flow&&level[e[i].y]==level[x]+1){
				if(d=getmaxflow(e[i].y,min(e[i].flow,flow-f))){
					f+=d;e[i].flow-=d;e[e[i].back].flow+=d;
					if(f==flow) return f;
				}
			}
		}
		if(!f) level[x]=-1;
		return f;
	}
	inline int dinic(){
		int ans=0,d;
		while(getlevel()){
			while(d=getmaxflow(s,inf)) ans+=d;
		}
		return ans;
	}
}
namespace zkww{
	inline bool getdis(){
		memset(vis,0,sizeof(vis));
		memset(dis,10,sizeof(dis));
		dis[t]=0;
		deque<int>q;q.push_back(t);
		while(!q.empty()){
			int tn=q.front();q.pop_front();
			for(int i=linkk[tn];i;i=e[i].next){
				if(dis[tn]-e[i].w<dis[e[i].y]&&e[e[i].back].flow){
					dis [e [i] .y] = dis [tn] -e [i] .w;
					if (! vis [e [i] .y]) {
						vis [e [i] .y] = 1;
						if(!q.empty()&&dis[e[i].y]<dis[q.front()]) q.push_front(e[i].y);
						else q.push_back(e[i].y);
					}
				}
			}
			vis [tn] = 0;
		}
		return dis[s]<168430090;
	}
	inline int getcost(int x,int flow){
        int f=0,d;vis[x]=1;
		if(x==t) return flow;
		for(int i=linkk[x];i;i=e[i].next){
			if(e[i].flow&&dis[e[i].y]==dis[x]-e[i].w&&!vis[e[i].y]){
				if(d=getcost(e[i].y,min(flow-f,e[i].flow))){
					f+=d;e[i].flow-=d;e[e[i].back].flow+=d;
					ans+=e[i].w*d;
					if(f==flow) return f;
				}
			}
		}
		return f;
	}
	inline void zkw(){
		while(getdis()){
			force[t]=1;
			while(vis[t]){
				memset(vis,0,sizeof(vis));
				getcost(s,inf);
			}
		}
	}
}
void build(){
	s=0;t=n;
	insert(s,1,k,0);
	for(int i=1;i<=m;i++){
		insert(x[i],y[i],inf,c[i]);
	}
}
int main(){
	using namespace dinicc;
	using namespace zkww;
	n=read();m=read();k=read();s=1;t=n;
	for(int i=1;i<=m;i++){
		x[i]=read();y[i]=read();f[i]=read();c[i]=read();
		insert(x[i],y[i],f[i],0);
	}
	ans=dinic();
	cout<<ans<<' ';
	build();ans=0;
	zkw ();
	cout<<ans<<endl;
	return 0;
}