网络流 -- Power Network（最大流算法）

</pre><div lang="en-US" class="ptt">Power Network</div><div class="plm"><table align="center"><tbody><tr><td><strong>Time Limit:</strong> 2000MS</td><td width="10"> </td><td><strong>Memory Limit:</strong> 32768K</td></tr><tr><td><strong>Total Submissions:</strong> 15399</td><td width="10"> </td><td><strong>Accepted:</strong> 8201</td></tr></tbody></table></div><p class="pst">Description</p><div lang="en-US" class="ptx">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<sub>max</sub>(u) of power, may consume an amount 0 <= c(u) <= min(s(u),c<sub>max</sub>(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<sub>max</sub>(u,v) of power delivered by u to v. Let Con=Σ<sub>u</sub>c(u) be the power consumed in the net. The problem is to compute the maximum value of Con. <center><img src="http://poj.org/images/1459_1.jpg" alt="" /></center>An example is in figure 1. The label x/y of power station u shows that p(u)=x and p<sub>max</sub>(u)=y. The label x/y of consumer u shows that c(u)=x and c<sub>max</sub>(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and l<sub>max</sub>(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. </div><p class="pst">Input</p><div lang="en-US" class="ptx">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<sub>max</sub>(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<sub>max</sub>(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<sub>max</sub>(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.</div><p class="pst">Output</p><div lang="en-US" class="ptx">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.</div><p class="pst">Sample Input</p><pre class="sio">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.

Source

Southeastern Europe 2003

 

 

直接套用最大流的模板的，主要是建图的过程。

输入分别为m个点，a个发电站，b个用户，n条边；接下去是n条边的信息(u,v)cost，cost表示边(u,v)的最大流量；a个发电站的信息(u)cost，cost表示发电站u能提供的最大流量；b个用户的信息(v)cost，cost表示每个用户v能接受的最大流量。
    典型的最大网络流中多源多汇的问题，在图中添加1个源点S和汇点T，将S和每个发电站相连，边的权值是发电站能提供的最大流量；将每个用户和T相连，边的权值是每个用户能接受的最大流量。从而转化成了一般的最大网络流问题，然后求解。

 

具体看程序吧，不解释.

<span style="font-family: Arial, Helvetica, sans-serif;">/*</span>

POJ 1459
*/
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<queue>
using namespace std;
const int MAXN=110;
const int INF=0x7fffffff;
int map[MAXN][MAXN],path[MAXN],flow[MAXN],start,end;
int n;//点的个数
queue<int>q;
int bfs()
{
    int i,t;
    while(!q.empty()) q.pop();//清空队列
    memset(path,-1,sizeof(path));
    path[start]=0;
    flow[start]=INF;
    q.push(start);
    while(!q.empty())
    {
        t=q.front();
        q.pop();
        if(t==end)  break;
        for(i=0;i<=n;i++)
        {
            if(i!=start&&path[i]==-1&&map[t][i])
            {
                flow[i]=flow[t]<map[t][i]?flow[t]:map[t][i];
                q.push(i);
                path[i]=t;
            }    
        }   
           
    }  
    if(path[end]==-1)  return -1;
        return flow[n];     
}       
int Edmonds_Karp()
{
    int max_flow=0,step,now,pre;
    while((step=bfs())!=-1)
    {
        max_flow+=step;
        now=end;
        while(now!=start)
        {
            pre=path[now];
            map[pre][now]-=step;
            map[now][pre]+=step;
            now=pre;
        }    
    }  
    return max_flow;  
}    
int main()
{
    int i,u,v,z,np,nc,m;
    while(scanf("%d%d%d%d",&n,&np,&nc,&m)!=EOF)
    {
        memset(map,0,sizeof(map));
        while(m--)
        {
            while(getchar()!='(');
            scanf("%d,%d)%d",&u,&v,&z);
            u++;v++;
            map[u][v]=z;
        }    
        while(np--)
        {
            while(getchar()!='(');
            scanf("%d)%d",&u,&z);
            u++;
            map[0][u]=z;
        }  
        while(nc--)
        {
            while(getchar()!='(');
            scanf("%d)%d",&u,&z);
            u++;
            map[u][n+1]=z;
        }  
        n++;
        start=0;end=n;
        printf("%d\n",Edmonds_Karp());    
    }    
    return 0;
}