poj1459 最大流之一般预流推进算法

Power Network

Time Limit: 2000MS		Memory Limit: 32768K
Total Submissions: 27505		Accepted: 14294

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

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.

题意：

给定一系列发电站，消费者，传送站，以及相应的发电量，消费量，传送量

求解这些站点中最大消费的电量可以是多少

题解：

运用网络流来求解，因此需要建立一个源点n+1和一个汇点n+2来构建图

一般预留推进算法思想：

         选择活跃顶点，通过它把一定的流量推进到它的邻接顶点，尽可能把每个点的余流减少为0

         算法基本框架：

                第一步：初始化源点余流为无穷大，源点距离标号为n

                第二步：从当前点出发，寻找活跃顶点，如果没有就结束程序

                第三步：如果到达汇点就把所推进的流加入的需求的答案中

                第四步：选取的活跃顶点中，如果有允许弧，则推进min（RE，EF）流

                第五步：推进完后判断当前的点是不是活跃顶点

                第六步：转第二步

#include
#include
#include
#include
using namespace std;

#define MAXN 110
#define INF 0x3f3f3f
int n,np,nc,m;
int s,t;//源点 汇点
int RE[MAXN][MAXN];//残留网络
queueque;
int H[MAXN];//高度
int EF[MAXN];//余流
int maxFlow;

void init()
{
    memset(RE,0,sizeof(RE));
    memset(H,0,sizeof(H));
    memset(EF,0,sizeof(EF));
    maxFlow=0;
    s=n+1,t=n+2;
    EF[s]=INF;
    EF[t]=-INF;
    H[s]=n;
}

void Push(int cur)
{
    for(int i=1;i<=t;i++){
        int temp=min(RE[cur][i],EF[cur]);
        
        if(temp>0&&(cur==s||H[cur]-H[i]==1)){
            //存在允许弧
            RE[cur][i]-=temp;
            RE[i][cur]+=temp;

            EF[cur]-=temp;
            EF[i]+=temp;
            
            if(i==t)
                maxFlow+=temp;
            else if(i!=s)
                que.push(i);
        }
    }
}

void Relabel(int cur)
{
    if(cur!=s&&cur!=t&&EF[cur]>0){
        H[cur]++;
        que.push(cur);
    }
}

void push_relabel()
{
    int cur;
    que.push(s);
    while(!que.empty()){
        cur = que.front();
        que.pop();
        Push(cur);
        //推进之后判断cur点是不是活跃顶点
        Relabel(cur);
    }
    printf("%d\n",maxFlow);
}


int main()
{
    //freopen("in.txt","r",stdin);
    while(scanf("%d%d%d%d",&n,&np,&nc,&m)!=EOF)
    {
        init();
        char ch;
        int u, v, val;
        for(int i=0;i