sgu 194 Reactor Cooling 题解【网络流】【上下界】

点击量：422

因为sgu搬到了CF上，所以附上题目地址

无源汇有上下界的可行流问题。

目录

Description

The terrorist group leaded by a well known international terrorist Ben Bladen is buliding a nuclear reactor to produce plutonium for the nuclear bomb they are planning to create. Being the wicked computer genius of this group, you are responsible for developing the cooling system for the reactor.

The cooling system of the reactor consists of the number of pipes that special cooling liquid flows by. Pipes are connected at special points, called nodes, each pipe has the starting node and the end point. The liquid must flow by the pipe from its start point to its end point and not in the opposite direction.

Let the nodes be numbered from $ 1$ to $ N$. The cooling system must be designed so that the liquid is circulating by the pipes and the amount of the liquid coming to each node (in the unit of time) is equal to the amount of liquid leaving the node. That is, if we designate the amount of liquid going by the pipe from $ i$-th node to $ j$-th as $ f_{ij}$, (put $ f_{ij}=0$ if there is no pipe from node $ i$ to node $ j$), for each $ i$ the following condition must hold:

$$ sum(j=1..N, f_{ij}) = sum(j=1..N, f_{ji}) $$

Each pipe has some finite capacity, therefore for each $ i$ and $ j$ connected by the pipe must be $ f_{ij}\le c_{ij}$ where $ c_{ij}$ is the capacity of the pipe. To provide sufficient cooling, the amount of the liquid flowing by the pipe going from $ i$-th to $ j$-th nodes must be at least $ l_{ij}$, thus it must be $ f_{ij}\ge l_{ij}$.

Given $ c_{ij}$ and $ l_{ij}$ for all pipes, find the amount $ f_{ij}$, satisfying the conditions specified above.

Input

The first line of the input file contains the number $ N (1 ≤ N ≤ 200)$ – the number of nodes and and $ M$ – the number of pipes. The following $ M$ lines contain four integer number each – $ i, j, l_{ij}$ and $ c_{ij}$ each. There is at most one pipe connecting any two nodes and $ 0\le l_{ij}\le c_{ij} \le 10^5$ for all pipes. No pipe connects a node to itself. If there is a pipe from $ i$-th node to $ j$-th, there is no pipe from $ j$-th node to $ i$-th.

Output

On the first line of the output file print YES if there is the way to carry out reactor cooling and NO if there is none. In the first case $ M$ integers must follow, $ k$-th number being the amount of liquid flowing by the $ k$-th pipe. Pipes are numbered as they are given in the input file.

Sample test(s)

Input

Test #1

4 6 
1 2 1 2 
2 3 1 2 
3 4 1 2 
4 1 1 2 
1 3 1 2 
4 2 1 2 

Test #2

4 6 
1 2 1 3 
2 3 1 3 
3 4 1 3 
4 1 1 3 
1 3 1 3 
4 2 1 3 

Output

Test #1

NO 

Test #2

YES 
1 
2 
3 
2 
1 
1 

Author:	Andrew Stankevich
Resource:	Petrozavodsk Winter Trainings 2003
Date:	2003-02-06

题意：

有$ n$个点和$ m$条管子，每个管子的流量有上下界。需要你找出一个方案，使得所有点的流入等于流出。即“没有一个点会凭空产生流量，也没有一个点会无端消耗流量”。

题解：

这个题的特点是无源汇和上下界。

对于无源汇，我们要找出一个可行的循环流。但是没有源汇怎么跑Dinic呢？

我们需要建立一个超级源点$ S$和超级汇点$ T$，来驱动这个循环流。同时要满足上下界的问题。

我们发现，当一条边合法时，它的流量一定在$ [l_{i,j},c_{i,j}]$范围内。因此我们帮这样的边跑掉那$ l_{i,j}$的流量，接下来它就是一条容量为$ c_{i,j}-l_{i,j}$的普通边了。

问题就在于怎么跑掉这个下界。前面提到了驱动，如果有边$ f_{i,j}$，那么它必须流入$ l_{i,j}$，并必须流出$ l_{i,j}$的流量。那么我们驱动它，就可以“帮”它跑出这么多流量。此时我们就可以连接$ f_{S,j}=l_{i,j}$，表示从源点连出来，这点流量不得不跑。也就是下图中绿色和橙色的边，它们称为附加边。

而此时$ f_{i,j}\le c_{i,j}-l_{i,j}$，那么这条边的实际流量是$ f_{i,j}+l_{i,j}$。

最后，当所有附加边全部满流时，说明所有上下界都达到了。具体而言，当绿色附加边满流时，说明这条边的下界已经达到了；当橙色附加边满流时，说明在没有超过其他边上界的情况下制造出了$ l_{i,j}$的循环流，其本质也是达到了这条边的下界。

这就是无源汇的可行流。

Code：

#include<cstdio>
#include<cstring>
int Min(int x,int y)
{
    return x<y?x:y;
}
struct edge
{
    int n,nxt,v,ori;
    edge(int n,int nxt,int v,int ori)
    {
        this->n=n;
        this->nxt=nxt;
        this->v=v;
        this->ori=ori;
    }
    edge(){}
}e[50000];
int head[300],ecnt=-1;
void add(int from,int to,int v,int ori)
{
    e[++ecnt]=edge(to,head[from],v,ori);
    head[from]=ecnt;
    e[++ecnt]=edge(from,head[to],0,0);
    head[to]=ecnt;
}
int d[300],q[300];
bool bfs()
{
    memset(d,0,sizeof(d));
    int l=0,r=0;
    q[++r]=233;
    d[233]=1;
    while(l<r)
    {
        int x=q[++l];
        for(int i=head[x];~i;i=e[i].nxt)
            if(e[i^1].v&&!d[e[i].n])
            {
                d[e[i].n]=d[x]+1;
                q[++r]=e[i].n;
            }
    }
    return d[0]>0;
}
int dinic(int x,int in)
{
    if(x==233)
        return in;
    int flow=in;    
    for(int i=head[x];flow&&~i;i=e[i].nxt)
        if(e[i].v&&d[e[i].n]==d[x]-1)
        {
            int tmp=dinic(e[i].n,Min(e[i].v,flow));
            if(!tmp)
                d[e[i].n]=0;
            e[i].v-=tmp;
            e[i^1].v+=tmp;
            flow-=tmp;
        }
    return in-flow;
}
int in[300],out[300];
int main()
{
    memset(head,-1,sizeof(head));
    int n,m,u,v,w,x;
    scanf("%d%d",&n,&m);
    int sum=0;
    for(int i=1;i<=m;++i)
    {
        scanf("%d%d%d%d",&u,&v,&w,&x);
        in[v]+=w;
        out[u]+=w;
        add(u,v,x-w,x);
        sum+=w;
    }
    for(int i=1;i<=n;++i)
    {
        add(0,i,in[i],0);
        add(i,233,out[i],0);
    }
    while(bfs())
    {
        int tmp;
        do
        {
            tmp=dinic(0,0x3fffffff);
            sum-=tmp;
        }while(tmp);
    }
    if(sum)
        puts("NO");
    else
    {
        puts("YES");
        for(int i=0;i<2*m;i+=2)
            printf("%d\n",e[i].ori-e[i].v);
    }
    return 0;
}