# POJ 1459 Power Network (最大流)

Power Network

 Time Limit: 2000MS Memory Limit: 32768K Total Submissions: 27622 Accepted: 14363

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) <= pmax(u) of power, may consume an amount 0 <= c(u) <= 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 <= l(u,v) <= 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.

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 <= 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 lmax(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 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 <= z <= 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

### AC代码

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<vector>
#include<queue>
using namespace std;

const int MAXN = 110;
const int MAXM = 110*110*2;
const int INF = 0x3f3f3f3f;

struct edge
{
int v,f;    //边终点与当前流量
int next;   //下一个兄弟位置
} edge[MAXM];

int n,np,nc,m;
int index;
int level[MAXN];

void init()
{
index =0;
}

void addedge(int u,int v,int f) //同时往图中添加原边与反向边
{
edge[index].v=v;
edge[index].f=f;
edge[index].v=u;
edge[index].f=0;
}

int bfs(int s,int t)    //利用bfs搜索当前残量图中是否存在s到t的路径
{
memset(level,0,sizeof(level));  //level为从源点到当前点的距离
level[s]=1;
queue<int>q;
q.push(s);
while(!q.empty())
{
int x=q.front();
q.pop();
if(x==t)return 1;
{
int v=edge[e].v,f=edge[e].f;
if(!level[v]&&f)
{
level[v]=level[x]+1;    //广度优先搜索下一层距离源点比当前多1
q.push(v);
}
}
}
return 0;
}

int dfs(int s,int maxf,int t)
{
if(s==t)return maxf;    //找到一条路径，返回当前路径中的最小流量
int ret=0;
{
int v=edge[e].v,f=edge[e].f;
if(level[s]+1==level[v]&&f) //如果新点可达并且标号为允许弧
{
int minn=min(maxf-ret,f);
f=dfs(v,minn,t);
edge[e].f-=f;
edge[e^1].f+=f;
ret+=f;
if(ret==maxf)return ret;
}
}
return ret;
}

int dinic(int s,int t)
{
int ans=0;
while(bfs(s,t)) //先确定是否存在路径从s到t
ans+=dfs(s,INF,t);  //
return ans;
}

int main()
{
while(~scanf("%d%d%d%d",&n,&np,&nc,&m)) //n作为源点，n+1作为汇点
{
init();
int ai,bi,ci;
for(int i=0; i<m; i++)  //ai到bi最大传输量为ci
{
while(getchar()!='(');
scanf("%d%*c%d%*c%d",&ai,&bi,&ci);
}
for(int i=0; i<np; i++) //ai最大发电量为bi
{
while(getchar()!='(');
scanf("%d%*c%d",&ai,&bi);
}