POJ训练计划1459_Power Network(网络流最大流/Dinic)

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解题报告

这题建模实在是好建，，，好贱，，，

给前向星给跪了，纯dinic的前向星竟然TLE，sad，，，回头看看优化，，，

矩阵跑过了，2A，sad，，，

/*************************************************************************
	> File Name:	PowerN.cpp
	> Author:		_nplus
	> Mail:	    jun18753370216@gmail.com
	> Time:		2014年07月19日 星期六 09时30分23秒
 ************************************************************************/

#include<cstdio>
#include<cmath>
#include<cstring>
#include<queue>
#include<iostream>
#include<algorithm>
#define inf 99999999
#define N 510
#define M N*N
using namespace std;
int edge[N][N],l[N],n,m,nc,np;
int bfs()
{
    queue<int >Q;
    memset(l,-1,sizeof(l));
    while(!Q.empty())
        Q.pop();
    l[n]=0;
    Q.push(n);
    while(!Q.empty())
    {
        int u=Q.front();
        Q.pop();
        for(int i=0; i<=n+1; i++)
        {
            if(edge[u][i]&&l[i]==-1)
            {
                l[i]=l[u]+1;
                Q.push(i);
            }
        }
    }
    if(l[n+1]>0)return 1;
    else return 0;
}
int dfs(int x,int f)
{
    if(x==n+1)return f;
    int a;
    for(int i=0; i<=n+1; i++)
    {
        if(edge[x][i]&&(l[i]==l[x]+1)&&(a=dfs(i,min(edge[x][i],f))))
        {
            edge[x][i]-=a;
            edge[i][x]+=a;
            return a;
        }
    }
    return 0;
}
int main()
{
    int i,j,u,v,w;
    while(~scanf("%d%d%d%d",&n,&np,&nc,&m))
    {
        memset(edge,0,sizeof(edge));
        for(i=0; i<m; i++)
        {
            while(getchar()!='(');
            scanf("%d,%d)%d",&u,&v,&w);
            edge[u][v]=w;
        }
        for(i=0; i<np; i++)
        {
            while(getchar()!='(');
            scanf("%d)%d",&v,&w);
            edge[n][v]=w;
        }
        for(i=0; i<nc; i++)
        {
            while(getchar()!='(');
            scanf("%d)%d",&u,&w);
            edge[u][n+1]=w;
        }
        int a,flow=0;
        while(bfs())
        {
            while(a=dfs(n,inf))
            {
                flow+=a;
            }
        }
        printf("%d\n",flow);
    }
}

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

Source

POJ训练计划1459_Power Network(网络流最大流/Dinic),布布扣,bubuko.com

POJ训练计划1459_Power Network(网络流最大流/Dinic)

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原文地址：http://blog.csdn.net/juncoder/article/details/37958293