POJ 1459 Power Network 多源点多汇点+最大流（EK算法）

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Power Network

Time Limit: 2000MS Memory Limit: 32768KTotal Submissions: 20469 Accepted: 10771

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)207 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

156

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

 

 

题的大体意思就是说一个电力网络有n个点，有np个发电站，nc个消耗点，其余的为中转站。m条电缆，中转站既不发电也不耗电。每条电缆都有一个最大容量。

思路就是设置一个超级源点和一个超级汇点将所有的源点和汇点分别放进去，然后直接EK算法。以下就是挫折码。

#include<stdio.h>#include<string.h>#include<iostream>#include<queue>#define INF 999999using namespace std;int cap[107][107],flow[107][107];//cap容量，flow流量int p[107],a[107];//p是父亲编号，a是残量int n,np,nc,m;queue<int>q;int EK(int s,int t){    int sum=0;    memset(p,0,sizeof(p));    while(1)    {        memset(a,0,sizeof(a));        q.push(s);        a[s]=INF;        while(!q.empty())        {            int u=q.front();            q.pop();            for(int i=0;i<=n+1;i++)            {                if(!a[i]&&cap[u][i]>flow[u][i])                {                    p[i]=u;                    a[i]=a[u]<cap[u][i]-flow[u][i]?a[u]:cap[u][i]-flow[u][i];                    q.push(i);                }            }        }        if(!a[t])break;        for(int i=t;i!=s;i=p[i])        {            flow[p[i]][i]+=a[t];            flow[i][p[i]]-=a[t];        }        sum+=a[t];    }    return sum;}int main(){    while(scanf("%d%d%d%d",&n,&np,&nc,&m)!=EOF)    {        memset(cap,0,sizeof(cap));        memset(flow,0,sizeof(flow));        char ch;        int a,b,u;        for(int i=0;i<m;i++)//边的信息        {            cin>>ch>>a>>ch>>b>>ch>>u;            cap[a+1][b+1]+=u;        }        for(int i=0;i<np;i++)//发电厂信息        {            cin>>ch>>a>>ch>>u;            cap[0][a+1]=u;        }        for(int i=0;i<nc;i++)//耗电用户的信息        {            cin>>ch>>a>>ch>>u;            cap[a+1][n+1]=u;        }        int ans=EK(0,n+1);        printf("%d\n",ans);    }    return 0;}