HDU3917 Road constructions 最大权闭合图 2011 Multi-University Training Contest 8 - Host by HUST

 

Road constructions

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 313 Accepted Submission(s): 113

Problem Description

N cities are required to connect with each other by a new transportation system. After several rounds of bidding, we have selected M constructions companies and
decided which section is assigned to which company, the associated cost and the direction of each road.

Due to the insufficiency of national fiscal revenue and special taxation system （the tax paid by each company pays is a fixed amount and tax payment occurs at the
beginning of the construction of the project） The government wishes to complete the project in several years and collects as much tax as possible to support the public
expense

For the restrictions of construction and engineering techniques, if a company is required to start the construction, then itself and its associated companies have to
complete all the tasks they commit (if company A constructs a road
from city 1 to city 2, company B constructs a road from city 2 to city 3, company C constructs a road from city 1 to city 3, we call
companies A and B are associated and other company pairs have no such relationship, pay attention, in this example and a are not associated, in other words,’
associated' is a directed relationship).
Now the question is what the maximum income the government can obtain in the first year is？

Input

There are multiple cases (no more than 50).
Each test case starts with a line, which contains 2 positive integers, n and m (1<=n<=1000, 1<=m<=5000).
The next line contains m integer which means the tax of each company.
The Third line has an integer k (1<=k<=3000)which indicates the number of the roads.
Then k lines fellow, each contains 4 integers, the start of the roads, the end of the road, the company is responsible for this road and the cost of the road.
The end of the input with two zero

Output

For each test case output the maximum income in a separate line, and if you can not get any income, please output 0.

Sample Input

   
   
   
   

    
    
    
    4 2
500 10
4
1 2 1 10
2 3 1 20
4 3 1 30
1 4 2 60
4 2
500 100
5
1 2 1 10
2 3 1 20
4 3 1 30
4 3 2 10
1 4 2 60
3 1
10
3
1 2 1 100
2 3 1 100
3 1 1 100
0 0
   
   
   
   

Sample Output

   
   
   
   

    
    
    
    440
470
0

    
    
    
    
 
     
 
      Hint 
     
for second test case, if you choose company 2 responsible ways, then you must choose the path of responsible company 1, but if you choose company 1, then you do not have to choose company 2. 
    
    
    
    
     
   
   
   
   

Source

2011 Multi-University Training Contest 8 - Host by HUST

Recommend

lcy

 

只比HDU3879多了一点：公司i是公司j的附属，那么i到j连边，容量为无限大。

 

#include<cstdio>
#include<cstring>
#include<vector>
#define N 8005
#define M 1000005
#define inf 999999999
#define min(a,b) ((a)<(b)?(a):(b))
using namespace std;

int n,m,s,t,num,adj[N],dis[N],q[N];
vector<int> st[1005],ed[1005];
struct edge
{
	int v,w,pre;
}e[M];
void insert(int u,int v,int w)
{
	e[num]=(edge){v,w,adj[u]};
	adj[u]=num++;
	e[num]=(edge){u,0,adj[v]};
	adj[v]=num++;
}
int bfs()
{
	int i,x,v,head=0,tail=0;
	memset(dis,0,sizeof(dis));
	dis[s]=1;
	q[tail++]=s;
	while(head<tail)
	{
		x=q[head++];
		for(i=adj[x];~i;i=e[i].pre)
			if(e[i].w&&!dis[v=e[i].v])
			{
				dis[v]=dis[x]+1;
				if(v==t)
					return 1;
				q[tail++]=v;
			}
	}
	return 0;
}
int dfs(int x,int limit)
{
	if(x==t)
		return limit;
	int i,v,tmp,cost=0;
	for(i=adj[x];~i&&cost<limit;i=e[i].pre)
		if(e[i].w&&dis[x]==dis[v=e[i].v]-1)
		{
			tmp=dfs(v,min(limit-cost,e[i].w));
			if(tmp)
			{
				e[i].w-=tmp;
				e[i^1].w+=tmp;
				cost+=tmp;
			}
			else
				dis[v]=-1;
		}
	return cost;
}
int Dinic()
{
	int ans=0;
	while(bfs())
		ans+=dfs(s,inf);
	return	ans;
}
int main()
{
	while(scanf("%d%d",&n,&m),n+m)
	{
		int i,j,k,u,v,c,w[5005]={0},sum=0;
		num=0;
		memset(adj,-1,sizeof(adj));
		s=0;
		t=1;
		for(i=1;i<=n;i++)
		{
			st[i].clear();
			ed[i].clear();
		}
		for(i=1;i<=m;i++)
		{
			scanf("%d",w+i);
			sum+=w[i];
			insert(i+1,t,w[i]);
		}
		scanf("%d",&k);
		for(i=1;i<=k;i++)
		{
			scanf("%d%d%d%d",&u,&v,&c,&w[0]);
			insert(s,i+m+1,w[0]);
			insert(i+m+1,c+1,inf);
			st[u].push_back(c);
			ed[v].push_back(c);
		}
		for(i=1;i<=n;i++)
		{
			for(j=0;j<ed[i].size();j++)
			{
				for(k=0;k<st[i].size();k++)
				{
					//printf("%d %d %d\n",i,ed[i][j],st[i][k]);
					insert(st[i][k]+1,ed[i][j]+1,inf);
				}
			}
		}
		printf("%d\n",sum-Dinic());
	}
}

 

由于选择了一个公司必须选择它的所有项目，所以可以把项目和公司合并为一个点。这样点和边都会减少。

#include<cstdio>
#include<cstring>
#include<vector>
#define N 5005
#define M 50005
#define inf 999999999
#define min(a,b) ((a)<(b)?(a):(b))
using namespace std;

int n,m,s,t,num,adj[N],dis[N],q[N];
vector<int> st[1005],ed[1005];
struct edge
{
	int v,w,pre;
}e[M];
void insert(int u,int v,int w)
{
	e[num]=(edge){v,w,adj[u]};
	adj[u]=num++;
	e[num]=(edge){u,0,adj[v]};
	adj[v]=num++;
}
int bfs()
{
	int i,x,v,head=0,tail=0;
	memset(dis,0,sizeof(dis));
	dis[s]=1;
	q[tail++]=s;
	while(head<tail)
	{
		x=q[head++];
		for(i=adj[x];~i;i=e[i].pre)
			if(e[i].w&&!dis[v=e[i].v])
			{
				dis[v]=dis[x]+1;
				if(v==t)
					return 1;
				q[tail++]=v;
			}
	}
	return 0;
}
int dfs(int x,int limit)
{
	if(x==t)
		return limit;
	int i,v,tmp,cost=0;
	for(i=adj[x];~i&&cost<limit;i=e[i].pre)
		if(e[i].w&&dis[x]==dis[v=e[i].v]-1)
		{
			tmp=dfs(v,min(limit-cost,e[i].w));
			if(tmp)
			{
				e[i].w-=tmp;
				e[i^1].w+=tmp;
				cost+=tmp;
			}
			else
				dis[v]=-1;
		}
	return cost;
}
int Dinic()
{
	int ans=0;
	while(bfs())
		ans+=dfs(s,inf);
	return	ans;
}
int main()
{
	while(scanf("%d%d",&n,&m),n+m)
	{
		int i,j,k,u,v,c,w[5005]={0},sum=0;
		num=0;
		memset(adj,-1,sizeof(adj));
		s=0;
		t=m+1;
		for(i=1;i<=n;i++)
		{
			st[i].clear();
			ed[i].clear();
		}
		for(i=1;i<=m;i++)
		{
			scanf("%d",w);
			sum+=w[0];
			insert(i,t,w[0]);
		}
		scanf("%d",&k);
		for(i=1;i<=k;i++)
		{
			scanf("%d%d%d%d",&u,&v,&c,w);
			w[c]+=w[0];
			st[u].push_back(c);
			ed[v].push_back(c);
		}
		for(i=1;i<=m;i++)
			insert(s,i,w[i]);
		for(i=1;i<=n;i++)
			for(j=0;j<ed[i].size();j++)
				for(k=0;k<st[i].size();k++)
					if(st[i][k]!=ed[i][j])
						insert(st[i][k],ed[i][j],inf);
		printf("%d\n",sum-Dinic());
	}
}