Poj 1459 Power Network【最大流Dinic---建图】

最新推荐文章于 2019-12-04 00:26:38 发布

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本文解析了一种电力网络模型，通过建立最大流问题求解电力网络中最大的电力消费值。介绍了如何构建网络模型，包括节点类型定义、边的容量限制，并详细说明了通过增加虚拟源点和汇点来解决该问题的方法。

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Power Network

Time Limit: 2000MS		Memory Limit: 32768K
Total Submissions: 26968		Accepted: 14020

Description

A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with an amount s(u) >= 0 of power, may produce an amount 0 <= p(u) <= p _max(u) of power, may consume an amount 0 <= c(u) <= min(s(u),c _max(u)) of power, and may deliver an amount d(u)=s(u)+p(u)-c(u) of power. The following restrictions apply: c(u)=0 for any power station, p(u)=0 for any consumer, and p(u)=c(u)=0 for any dispatcher. There is at most one power transport line (u,v) from a node u to a node v in the net; it transports an amount 0 <= l(u,v) <= l _max(u,v) of power delivered by u to v. Let Con=Σ _uc(u) be the power consumed in the net. The problem is to compute the maximum value of Con.

An example is in figure 1. The label x/y of power station u shows that p(u)=x and p _max(u)=y. The label x/y of consumer u shows that c(u)=x and c _max(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and l _max(u,v)=y. The power consumed is Con=6. Notice that there are other possible states of the network but the value of Con cannot exceed 6.

Input

There are several data sets in the input. Each data set encodes a power network. It starts with four integers: 0 <= n <= 100 (nodes), 0 <= np <= n (power stations), 0 <= nc <= n (consumers), and 0 <= m <= n^2 (power transport lines). Follow m data triplets (u,v)z, where u and v are node identifiers (starting from 0) and 0 <= z <= 1000 is the value of l _max(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 10000 is the value of p _max(u). The data set ends with nc doublets (u)z, where u is the identifier of a consumer and 0 <= z <= 10000 is the value of c _max(u). All input numbers are integers. Except the (u,v)z triplets and the (u)z doublets, which do not contain white spaces, white spaces can occur freely in input. Input data terminate with an end of file and are correct.

Output

For each data set from the input, the program prints on the standard output the maximum amount of power that can be consumed in the corresponding network. Each result has an integral value and is printed from the beginning of a separate line.

Sample Input

2 1 1 2 (0,1)20 (1,0)10 (0)15 (1)20
7 2 3 13 (0,0)1 (0,1)2 (0,2)5 (1,0)1 (1,2)8 (2,3)1 (2,4)7
         (3,5)2 (3,6)5 (4,2)7 (4,3)5 (4,5)1 (6,0)5
         (0)5 (1)2 (3)2 (4)1 (5)4

Sample Output

15
6

Hint

The sample input contains two data sets. The first data set encodes a network with 2 nodes, power station 0 with pmax(0)=15 and consumer 1 with cmax(1)=20, and 2 power transport lines with lmax(0,1)=20 and lmax(1,0)=10. The maximum value of Con is 15. The second data set encodes the network from figure 1.

Source

Southeastern Europe 2003

题目大意：一共有n个点，有np个生产点，有nc个消费点，有m条有向边，问最多消费者会购买多少单位的东西。接下来m个三元组，表示有从点u到点v最多能够运输w个单位的东西，再接下来np个二元组，表示生产点u能够生产w个单位的东西，同样，再接下来nc个二元组，表示消费点u能够买w个单位的东西。

思路：

1、首先对m条边，直接对从u到v建边。

2、然后对于np个生产点，用一个超级源点连接这些个生产点，其边权值为生产点最多能够生产的物品单位数量。接着对于nc个消费点，将其连入一个超级汇点，每条边权值为消费点能够购买的物品单位数量。

3、这时候源点S也有了，汇点T也有了，直接跑一遍最大流算法即可。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
#define INF 0x3f3f3f3f
struct edege
{
    int from;
    int to;
    int w;
    int next;
}e[1200000];
int head[120000];
int div[1200000];
int n,np,nc,m,cont,ss,tt;
void add(int from,int to,int w)
{
    e[cont].to=to;
    e[cont].w=w;
    e[cont].next=head[from];
    head[from]=cont++;
}
int makediv()
{
    memset(div,0,sizeof(div));
    div[ss]=1;
    queue<int >s;
    s.push(ss);
    while(!s.empty())
    {
        int u=s.front();
        if(u==tt)return 1;
        s.pop();
        for(int i=head[u];i!=-1;i=e[i].next)
        {
            int v=e[i].to;
            int w=e[i].w;
            if(w&&div[v]==0)
            {
                div[v]=div[u]+1;
                s.push(v);
            }
        }
    }
    return 0;
}
int Dfs(int u,int maxflow,int tt)
{
    if(tt==u)return maxflow;
    int ret=0;
    for(int i=head[u];i!=-1;i=e[i].next)
    {
        int v=e[i].to;
        int w=e[i].w;
        if(w&&div[v]==div[u]+1)
        {
            int f=Dfs(v,min(maxflow-ret,w),tt);
            e[i].w-=f;
            e[i^1].w+=f;
            ret+=f;
            if(ret==maxflow)return ret;
        }
    }
    return ret;
}
void Dinic()
{
    int ans=0;
    while(makediv()==1)
    {
        ans+=Dfs(ss,INF,tt);
    }
    printf("%d\n",ans);
}
int main()
{
    while(~scanf("%d%d%d%d",&n,&np,&nc,&m))
    {
        cont=0;
        ss=n+1;
        tt=ss+1;
        memset(head,-1,sizeof(head));
        for(int i=0;i<m;i++)
        {
            int x,y,w;
            scanf(" (%d,%d)%d",&x,&y,&w);
            add(x,y,w);
            add(y,x,0);
        }
        for(int i=0;i<np;i++)
        {
            int v,w;
            scanf(" (%d)%d",&v,&w);
            add(ss,v,w);
            add(v,ss,0);
        }
        for(int i=0;i<nc;i++)
        {
            int u,w;
            scanf(" (%d)%d",&u,&w);
            add(u,tt,w);
            add(tt,u,0);
        }
        Dinic();
    }
}