poj1459 Power Network [最大流]

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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) \ge 0 $ of power, may produce an amount $ 0 \le p(u) \le pmax(u) of power $ , may consume an amount $ 0 \le c(u) \le min(s(u),cmax(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 \le l(u,v) \le lmax(u,v) of power $ delivered by u to v. Let Con=Σuc(u) be the power consumed in the net. The problem is to compute the maximum value of Con.

题目

An example is in figure 1. The label x/y of power station u shows that p(u)=x and pmax(u)=y. The label x/y of consumer u shows that c(u)=x and cmax(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and lmax(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 \le n \le 100 (nodes) $ , $ 0 \le np \le n $(power stations),$ 0 \le nc \le n $(consumers), and $ 0 \le m \le n^2 $ (power transport lines). Follow m data triplets (u,v)z, where u and v are node identifiers (starting from 0) and $ 0 \le z \le 1000 $ is the value of lmax(u,v). Follow np doublets (u)z, where u is the identifier of a power station and $ 0 \le z \le 10000 $ is the value of pmax(u). The data set ends with nc doublets (u)z, where u is the identifier of a consumer and $ 0 \le z \le 10000 $ is the value of cmax(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.


Solution

最大流,建图后用dinic求解.建图方法:

  1. 前m组传输电线,$ u\rightarrow v $ 权值为w
  2. 之后np组电站,$ S(超级源)\rightarrow x $ 权值为w,等效于源点提供w的电能
  3. 之后nc组消费者,$ x\rightarrow T(超级汇) $ 权值为w等效于向汇点提供w的电能
    样例

Code

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#define maxn 220
#define maxe 30010
#define INF 0x3f3f3f3f
#define DEBUG
using namespace std;
int n,np,nc,m,S,T;
int cnt,head[maxn];
int dis[maxn];

struct edge
{
    int to,next,val;
}e[maxe];

inline int read()
{
    char ch;
    int read=0,sign=1;
    do
        ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-');
    if(ch=='-') sign=-1,ch=getchar();
    while(ch>='0'&&ch<='9')
    {
        read=read*10+ch-'0';
        ch=getchar();
    }
    return sign*read;
}

inline void ins(int u,int v,int w)
{
    ++cnt;
    e[cnt].to=v;
    e[cnt].val=w;
    e[cnt].next=head[u];
    head[u]=cnt;
}

inline void insert(int u,int v,int w)
{ ins(u,v,w);ins(v,u,0); }

void init()
{
    np=read(),nc=read(),m=read(),S=0,T=n+1,cnt=1;
    memset(e,0,sizeof(e));
    memset(head,0,sizeof(head));
}

void build_graph()
{
    for(int i=1; i<=m; ++i)
    {
        int u,v,w;
        u=read()+1,v=read()+1,w=read();
        insert(u,v,w); 
    }
    for(int i=1; i<=np; ++i)
    {
        int x,w;
        x=read()+1,w=read();
        insert(S,x,w);
    }
    for(int i=1; i<=nc; ++i)
    {
        int x,w;
        x=read()+1,w=read();
        insert(x,T,w);
    }
}

bool bfs()
{
    memset(dis,-1,sizeof(dis));
    queue<int> q;
    q.push(0);
    dis[0]=0;
    while(!q.empty())
    {
        int u=q.front();q.pop();
        int i=head[u];
        while(i)
        {
            if( e[i].val && dis[e[i].to]<0 )
            {
                dis[e[i].to]=dis[u]+1;
                q.push(e[i].to);
            }
            i=e[i].next;
        }
    }
    if(dis[T]==-1) return 0;
    return 1;
} 

int dfs(int x,int flow)
{
    if(x==T) return flow;
    int i=head[x],used=0,w;
    while(i)
    {
        if(e[i].val && dis[e[i].to]==dis[x]+1)
        {
            w=flow-used;
            w=dfs(e[i].to,min(e[i].val,w));
            e[i].val-=w;
            e[i^1].val+=w;
            used+=w;
            if(used==flow) return flow;
        }
        i=e[i].next;
    } 
    if(!used) dis[x]=-1;
    return used;
} 

int dinic()
{
    int ans=0;
    while(bfs()) ans+=dfs(S,INF);
    return ans;
}

int main()
{
    while(scanf("%d",&n)==1)
    {
        init();
        build_graph();
        printf("%d\n",dinic());
    }
    return 0;
}

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