poj1459 Power Network(最大流)

Power Network

Time Limit: 2000MS		Memory Limit: 32768K
Total Submissions: 29893		Accepted: 15475

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.

 

分析：最大流的裸题，题目不难理解，建图的时候选0做超级源点，每个点的下标加1，

0到电站连边的容量为电站的最大发电量p_max ，consumer拆成两个点u和u+Nc，它们之间

边的容量即为consumer的最大接受电量c_max ，然后u+Nc与超级汇点T连边，容量为INF。

最后用Dinic算出最大流，注意题目的输入形式就好了。

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#define INF 999999999
using namespace std;
int N,Np,Nc,M,T;//M条边,Np个电站,Nc个consumer 
int map[400][400],dis[402];
char s[100];
int bfs()
{
    memset(dis,-1,sizeof(dis));
    dis[0]=0;
    queue<int> q;
    q.push(0);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int i=0;i<=T;i++)
        if(dis[i]==-1&&map[u][i]>0)
        {
            dis[i]=dis[u]+1;
            q.push(i);
        }
    }
    if(dis[T]>0) return 1;
    return 0;
}

int dfs(int cur,int m)
{
    if(cur==T) return m;
    int f,res=0;
    for(int i=0;i<=T;i++)
    {
        if(dis[i]==dis[cur]+1&&map[cur][i]>0&&(f=dfs(i,min(m,map[cur][i]))))
        {
            map[cur][i]-=f;
            map[i][cur]+=f;
            res+=f;
            m-=f;
            if(!m) break;
        }
    }
    if(res) return res;
    dis[cur]=-1;
    return 0;
}

int main()
{
    while(scanf("%d%d%d%d",&N,&Np,&Nc,&M)!=-1)
    {
        memset(map,0,sizeof(map));
        //每个点下标加1
        T=2*N+1+Nc;//T为汇点(数据量小,随便找个大一点的做汇点就行),源点为0,consumer拆为两个点限流 
        for(int i=0;i<M;i++)//M条边 
        {
            int u=0,v=0,c=0;
            scanf("%s",s);
            int j=1;
            while(s[j]!=',')
            {
                u=u*10+s[j]-'0';
                j++;
            }
            j++;
            while(s[j]!=')')
            {
                v=v*10+s[j]-'0';
                j++;
            }
            j++;
            while(s[j])
            {
                c=c*10+s[j]-'0';
                j++;
            }
            map[u+1][v+1]+=c;
        }
        for(int i=0;i<Np;i++)//Np个电站 
        {
            int u=0,p=0,j=1;
            scanf("%s",s);
            while(s[j]!=')')
            {
                u=u*10+s[j]-'0';
                j++;
            }
            j++;
            while(s[j])
            {
                p=p*10+s[j]-'0';
                j++;
            }
            map[0][u+1]+=p;//源点->电站
        }
        
        for(int i=0;i<Nc;i++)//Nc个consumer 
        {
            scanf("%s",s);
            int u=0,c=0,j=1;
            while(s[j]!=')')
            {
                u=u*10+s[j]-'0';
                j++;
            }
            j++;
            while(s[j])
            {
                c=c*10+s[j]-'0';
                j++;
            }
            map[u+1][u+1+N]+=c;//consumer拆为两个点限流 
            map[u+1+N][T]=INF;//consumer->汇点流量为INF
        }
        
        int ans=0,res;
        while(bfs())
            while(res=dfs(0,INF))
                ans+=res;
        printf("%d\n",ans);
    }
    return 0;
}