poj 1459 Power Network （dinic）

标签：des   style   http   color   os   io   strong   for   

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

Output

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

#include"stdio.h"
#include"string.h"
#include"queue"
#include"algorithm"
using namespace std;
#define N 205
#define inf 100000000
int g[N][N],dis[N];
int min(int a,int b)
{
    return a<b?a:b;
}
int spfa(int s,int t)
{
    int i;
    queue<int>q;
    q.push(s);
    memset(dis,0,sizeof(dis));
    dis[s]=1;
    while(!q.empty())
    {
        s=q.front();
        q.pop();
        for(i=0;i<=t;i++)
        {
            if(!dis[i]&&g[s][i])
            {
                dis[i]=dis[s]+1;
                if(i==t)
                    return 1;
                q.push(i);
            }
        }
    }
    return 0;
}
int dfs(int s,int t,int lim)
{
    int i,cost=0,tmp;
    if(s==t)
        return lim;
    for(i=0;i<=t;i++)
    {
        if(g[s][i]&&dis[i]==dis[s]+1)
        {
            tmp=dfs(i,t,min(lim-cost,g[s][i]));
            if(tmp>0)
            {
                g[s][i]-=tmp;
                g[i][s]+=tmp;
                cost+=tmp;
                if(cost==lim)
                    break;
            }
            else
                dis[i]=-1;
        }
    }
    return cost;
}
int dinic(int s,int t)
{
    int ans=0;
    while(spfa(s,t))
        ans+=dfs(s,t,inf);
    return ans;
}
int main()
{
    int n,np,nc,m,u,v,d,i;
    char str[30];
    while(scanf("%d%d%d%d",&n,&np,&nc,&m)!=-1)
    {
        memset(g,0,sizeof(g));
        while(m--)
        {
            scanf("%s",str);
            u=v=d=0;
            for(i=1;str[i]!='\0';i++)
            {
                if(str[i]>='0'&&str[i]<='9')
                    u=u*10+str[i]-'0';
                else break;
            }
            for(i++;str[i]!='\0';i++)
            {
                if(str[i]>='0'&&str[i]<='9')
                    v=v*10+str[i]-'0';
                else break;
            }
            for(i++;str[i]!='\0';i++)
            {
                if(str[i]>='0'&&str[i]<='9')
                    d=d*10+str[i]-'0';
                else break;
            }
            g[u][v]+=d;
        }
        while(np--)
        {
            scanf("%s",str);
            u=d=0;
            for(i=1;str[i]!='\0';i++)
            {
                if(str[i]>='0'&&str[i]<='9')
                    u=u*10+str[i]-'0';
                else break;
            }
            for(i++;str[i]!='\0';i++)
            {
                if(str[i]>='0'&&str[i]<='9')
                    d=d*10+str[i]-'0';
                else break;
            }
            g[n+1][u]+=d;          //超级源点
        }
        while(nc--)
        {
            scanf("%s",str);
            u=d=0;
            for(i=1;str[i]!='\0';i++)
            {
                if(str[i]>='0'&&str[i]<='9')
                    u=u*10+str[i]-'0';
                else break;
            }
            for(i++;str[i]!='\0';i++)
            {
                if(str[i]>='0'&&str[i]<='9')
                    d=d*10+str[i]-'0';
                else break;
            }
            g[u][n+2]+=d;          //超级汇点
        }
        int ans=dinic(n+1,n+2);
        printf("%d\n",ans);
    }
    return 0;
}

poj 1459 Power Network （dinic）,布布扣,bubuko.com

poj 1459 Power Network （dinic）

标签：des   style   http   color   os   io   strong   for   

原文地址：http://blog.csdn.net/u011721440/article/details/38611197