CUGB图论专场2：D - Power Network 最大流EK算法

D - Power Network

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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.

思路：刚开始不理解EK算法中正向与反向边中的加减，所以查找了好多资料，有了前一篇博客的转载才理解其中的意思，所以才完成了这道题，可以说搞了一天了，哎呀，刚才又因为输入的问题卡了好久，晕死……

因为所以的电站都是源点，所有的消费者都是汇点，所以新建一个源点把所有的电站连起来，再新建一个汇点把所有的消费者连起来就行了，我把有的点加1，就是1到n，然后0点为源点，n+1点为汇点，用EK算法就可以算出来了。

EK算法是比较简单的最大流算法，比较好理解，代码比较少。现在一步一步学习最大流算法，今天学习了最大流的所有算法，都理解了其中的意味，不过还没有敲代码，有时间再一个一个算法的敲了。

EK算法中，正向减去，因为流题加上就相当于容量减去，其差是一样的，所以可以直接把容量减去就行了。但是为何其反向边要加上呢，因为如果没有反向边的话，增广路是不能回退的，即可能还存在别的增广路使最大流更加大。比如：1，2，3，4结点之间容量都是1，若1，2，3，4是增广路，然后就不存在增广路了，则最大流为1；但是这个最大流是错的，因为可以过1，2，4这条增广路以及1，3，4这条增广路使得最大流为2.如果加入反向边，那么寻找第一次增广路即1，2，3，4的时候，反向边（3，2)加了一个数，那么就不为0，此时最大流为1；则可以寻找到第二条增广路1，3，2，4，此时最大流为2正确…………（详细的在前一篇博客中有转载的详解）

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <list>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <set>
#define PI acos(-1.0)
#define mem(a,b) memset(a,b,sizeof(a))
#define sca(a) scanf("%d",&a)
#define pri(a) printf("%d\n",a)
#define MM 10002
#define MN 105
#define INF 168430090
using namespace std;
typedef long long ll;
int n,m,k,t,ans,pre[MN],vis[MN],Map[MN][MN];
int bfs()
{
    mem(vis,0); queue<int>q; q.push(0); vis[0]=1;
    while(!q.empty())  //找增广路
    {
        int u=q.front(); q.pop();
        for(int v=0;v<=n+1;v++)
        {
            if(!vis[v]&&Map[u][v])
            {
                pre[v]=u;  //记录前向结点
                if(v==n+1) return 1;
                q.push(v); vis[v] = 1;
            }
        }
    }
    return 0;
}
void change()
{
    int i,sum=INF;
    for(i=n+1; i != 0; i = pre[i])
        sum = min(sum,Map[pre[i]][i]);
    for(i=n+1; i != 0; i = pre[i])
    {
        Map[pre[i]][i]-=sum; //正向边减
        Map[i][pre[i]]+=sum; //反向边加
    }
    ans+=sum;
}
int main()
{
    while(~scanf("%d%d%d%d",&n,&m,&k,&t))
    {
        mem(Map,0); ans=0;
        int u,v,w;
        while(t--)
        {
            while(getchar()!='(');
            scanf("%d,%d)%d",&u,&v,&w);
            Map[u+1][v+1]+=w;
        }
        while(m--)
        {
            while(getchar()!='(');
            scanf("%d)%d",&u,&w);
            Map[0][u+1]=w;  //新建源点0
        }
        while(k--)
        {
            while(getchar()!='(');
            scanf("%d)%d",&u,&w);
            Map[u+1][n+1]=w; //新建汇点n+1
        }
        while(bfs()) change();
        pri(ans);
    }
    return 0;
}