Power Network 网络流入门题)

这个题 一看到字符串处理我就烦啊烦== 在小伙伴的帮助下可算是知道怎么改对了 万恶的格式（你怎么不说自己学的不好-->_-->）

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

#include <stdio.h>
#include <iostream>
using namespace std;
const int oo=1e9;
const int mm=111111;
const int mn=999;
int node ,scr,dest,edge;
int ver[mm],flow[mm],next[mm];
int head[mn],work[mn],dis[mn],q[mn];
void prepare(int _node,int _scr,int _dest)
{
    node=_node,scr=_scr,dest=_dest;
    for(int i=0; i<node; ++i)
        head[i]=-1;
    edge=0;
}
void addedge(int u,int v,int c)
{
    ver[edge]=v,flow[edge]=c,next[edge]=head[u],head[u]=edge++;
    ver[edge]=u,flow[edge]=0,next[edge]=head[v],head[v]=edge++;
}
bool Dinic_bfs()
{
    int i,u,v,l,r=0;
    for(i=0; i<node; i++)
        dis[i]=-1;
    dis[q[r++]=scr]=0;
    for(l=0; l<r; ++l)
    {
        for(i=head[u=q[l]]; i>=0; i=next[i])
        {
            if(flow[i]&&dis[v=ver[i]]<0)
            {
                dis[q[r++]=v]=dis[u]+1;
                if(v==dest)
                    return 1;
            }
        }
    }
    return 0;
}
int Dinic_dfs(int u,int exp)
{
    if(u==dest)
        return exp;
    for(int &i=work[u],v,tmp; i>=0; i=next[i])
        if(flow[i]&&dis[v=ver[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0)
        {
            flow[i]-=tmp;
            flow[i^1]+=tmp;
            return tmp;
        }
    return 0;
}
int Dinic_flow()
{
    int i,ret=0,delta;
    while(Dinic_bfs())
    {
        for(i=0; i<node; i++)
            work[i]=head[i];
        while(delta=Dinic_dfs(scr,oo))
            ret+=delta;
    }
    return ret;
}
int main()
{
    //freopen("cin.txt","r",stdin);
    int n,np,nc,m,z,u,v;
    char str;
    while(~scanf("%d%d%d%d",&n,&np,&nc,&m))
    {
       // printf("%d %d %d %d\n",n,np,nc,m);
        prepare(n+3,0,n+1);
        for(int i=0;i<m;i++)
        {
            while(getchar()!='(');
            scanf("%d,%d)%d",&u,&v,&z);
            // printf("%d    %d    %d\n",u,v,z);
             addedge(u+1,v+1,z);
        }
        for(int i=0;i<np;i++)
        {
            while(getchar()!='(');
            scanf("%d)%d",&u,&z);
            //printf("%d  %d\n",u,z);
             addedge(0,u+1,z);
        }
        for(int i=0;i<nc;i++)
        {
            while(getchar()!='(');
            scanf("%d)%d",&u,&z);
           // printf("%d      %d\n",u,z);
            addedge(u+1,n+1,z);
        }
        printf("%d\n",Dinic_flow());
    }
    return 0;
}