# POJ 1459 Power Network / HIT 1228 Power Network / UVAlive 2760 Power Network / ZOJ 1734 Power Network / FZU 1161 （网络流，最大流）

### 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

15
6

## 代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxN=300;
const int maxM=maxN*maxN*2;
const int inf=2147483647;

class Edge
{
public:
int u,v,flow;
};

int n,np,nc;//总机械数，供能机械数，收能机械数
int m;//能量管道数
int cnt=-1;
int Next[maxM];
Edge E[maxM];
int depth[maxN];
int cur[maxN];
int Q[maxN];

bool bfs();
int dfs(int u,int flow);

int main()
{
while (cin>>n>>np>>nc>>m)//多组数据
{
cnt=-1;//每次先要清空
while (m--)//读入边
{
int u,v,w;
while (getchar()!='(');//注意这里读入时要特殊处理一下
scanf("%d,%d)%d",&u,&v,&w);
}
while (np--)//读入供能机械
{
int u,maxflow;
while (getchar()!='(');
scanf("%d)%d",&u,&maxflow);
}
while (nc--)//读入收能机械
{
int u,maxflow;
while (getchar()!='(');
scanf("%d)%d",&u,&maxflow);
}
/*for (int i=0;i<=n+1;i++)
{
cout<<i<<" "<<E[j].v<<" "<<E[j].flow<<endl;
cout<<endl;
}
//*/
int Ans=0;//求解最大流，Dinic算法
while (bfs())
{
for (int i=0;i<=n+1;i++)
while (int di=dfs(0,inf))
Ans+=di;
}
printf("%d\n",Ans);
}
return 0;
}

{
cnt++;
E[cnt].u=u;
E[cnt].v=v;
E[cnt].flow=flow;

cnt++;
E[cnt].u=v;
E[cnt].v=u;
E[cnt].flow=0;
}

bool bfs()
{
memset(depth,-1,sizeof(depth));
int h=1,t=0;
depth[0]=1;
Q[1]=0;
do
{
t++;
int u=Q[t];
{
int v=E[i].v;
if ((E[i].flow>0)&&(depth[v]==-1))
{
depth[v]=depth[u]+1;
h++;
Q[h]=v;
}
}
}
while (t!=h);
if (depth[n+1]==-1)
return 0;
return 1;
}

int dfs(int u,int flow)
{
if (u==n+1)
return flow;
for (int &i=cur[u];i!=-1;i=Next[i])
{
int v=E[i].v;
if ((depth[v]==depth[u]+1)&&(E[i].flow>0))
{
int di=dfs(v,min(flow,E[i].flow));
if (di>0)
{
E[i].flow-=di;
E[i^1].flow+=di;
return di;
}
}
}
return 0;
}


posted @ 2017-08-16 20:10  SYCstudio  阅读(376)  评论(2编辑  收藏  举报