poj 1459 Power Network

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                                         ***Power Network***

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

题目大意：
给n个发电站，给np个消耗站，再给nc个转发点。
发电站只发电，消耗站只消耗电，转发点只是转发电，再给m个传送线的传电能力。
问你消耗站能获得的最多电是多少。
解题思路：
网络流：
增加一个源点和一个汇点就ok

上代码：

/*
Date : 2015-8-21 下午

Author : ITAK

Motto :

今日的我要超越昨日的我，明日的我要胜过今日的我；
以创作出更好的代码为目标，不断地超越自己。
*/
#include <iostream>
#include <cstdio>

using namespace std;
///oo表示无穷大
const int oo = 1e9+5;
///mm表示边的最大数量，因为要双向建边
const int mm = 111111;
///点的最大数量
const int mn = 1000;
///node:节点数,src:源点,dest:汇点,edge:边数
int node, src, dest, edge;
///ver:边指向的结点,flow:边的流量,next:链表的下一条边
int ver[mm], flow[mm], next[mm];
///head:节点的链表头,work:用于算法中的临时链表头,dis:距离
int head[mn], work[mn], dis[mn], q[mn];

///初始化
void Init(int _node, int _src, int _dest)
{
    node = _node, src = _src, dest = _dest;
    for(int i=0; i<node; i++)
        head[i] = -1;
    edge = 0;
}

///增加边
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++;
}

///广搜计算出每个点与源点的最短距离，如果不能到达汇点说明算法结束
bool Dinic_bfs()
{
    int i, u, v, l, r = 0;
    for(i=0; i<node; i++)
        dis[i] = -1;
    dis[q[r++]=src] = 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)
            {
                ///这条边必须有剩余流量
                dis[q[r++]=v] = dis[u] + 1;
                if(v == dest)
                    return 1;
            }
    return 0;
}

///寻找可行流的增广路算法，按节点的距离来找，加快速度
int Dinic_dfs(int u, int exp)
{
    if(u == dest)
        return exp;
    ///work 是临时链表头，这里用 i 引用它，这样寻找过的边不再寻找*
    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)
        {
            ///正反向边容量改变
            flow[i] -= tmp;
            flow[i^1] += tmp;
            ///异或运算，相当于i(偶数)，与i+1互换,奇数i与i-1互换。2^1=3,3^1=2;
            return tmp;
        }
    }

    return 0;
}

///求最大流，直到没有可行流
int Dinic_flow()
{
    int i, ret=0, data;
    while(Dinic_bfs())
    {
        for(i=0; i<node; i++)
            work[i] = head[i];
        while(data = Dinic_dfs(src, oo))
            ret += data;
    }

    return ret;
}
int main()
{
    int m, n, np, nc;
    int u, v, w;
    while(cin>>n>>np>>nc>>m)
    {
        Init(n+np+2, n, n+1);
        while(m--)
        {
            scanf(" (%d,%d)%d",&u,&v,&w);
            addedge(u, v, w);
        }
        while(np--)
        {
            scanf(" (%d)%d",&u,&w);
            addedge(src, u, w);
        }
        while(nc--)
        {
            scanf(" (%d)%d",&u,&w);
            addedge(u, dest, w);
        }
        printf("%d\n",Dinic_flow());
    }
    return 0;
}