POJ 1459 Power Network (最大流)

Power Network

 

Time Limit: 2000MS Memory Limit: 32768K
Total Submissions: 27622 Accepted: 14363

 

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) <= pmax(u) of power, may consume an amount 0 <= c(u) <= min(s(u),cmax(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) <= lmax(u,v) of power delivered by u to v. Let Con=Σuc(u) be the power consumed in the net. The problem is to compute the maximum value of Con.

1459_1
An example is in figure 1. The label x/y of power station u shows that p(u)=x and pmax(u)=y. The label x/y of consumer u shows that c(u)=x and cmax(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and lmax(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 lmax(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 10000 is the value of pmax(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 cmax(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

 

题意

这道题题目好长,好难懂哎!看了好久的题目。

先附上翻译(网上找的)

题目描述

一个电网包含一些结点(电站、消费者、调度站),这些结点通过电线连接。每个结点 u 可能被供给 s(u) 的电能, s(u)≥0 ,同时也可能产生 p(u) 的电能, 0≤p(u)≤pmax(u) ,站点 u 还有可能 消费 c(u) 电能, 0≤c(u)≤min( s(u),cmax(u)) ,可能传输 d(u) 的电能, d(u) = s(u) + p(u) - c(u) 。

以上这些量存在以下限制关系:

对每个电站, c(u) = 0 。对每个消费者, p(u) = 0 。对每个调度站, p(u) = c(u) = 0 。

在电网中两个结点 u 和 v 之间最多有一条电线连接。从结点 u 到结点 v 传输 L(u,v) 的电能, 0≤L(u,v)≤Lmax(u,v) 。定义 Con 为 c(u) 的总和,表示电网中消费电能的总和。本题的目的是求 Con 的最大值。

在图(a)中,电站结点 u 的标记” x/y ”代表 p(u) = x 、 pmax(u) = y 。消费者结点 u 的标记” x/y ”代表 c(u) = x 、 cmax(u) = y 。每条电线所对应的边 (u,v) ,其标记” x/y ” 代表 L(u,v) = x 、 Lmax(u,v) = y 。

在图(b)中,消费的最大电能 Con = 6 ,图(a)列出了在此状态下各 个站点的 s(u) 、 p(u) 、 c(u) 和 d(u) 。注意,如图(b)所示的电网中,电能的流动还存在其他状态,但 消费的电能总和不超过6。

输入描述

输入文件中包含多个测试数据,每个测试数据描述了一个电网。每个测试数据的第1 行为4 个整数: n,np,nc,m ,其中, 0≤n≤100 ,代表结点数目; 0≤np≤n ,代表电站数目; 0≤nc≤n , 代表消费者数目; 0≤m≤n2 ,代表传输电线的数目。接下来有 m 个三元组, (u,v)z ,其中 u 和 v 为结点序号(结点序号从0开始计起), 0≤z≤1000 ,代表 Lmax(u,v) 的值。接下来有 np 个二元 组, (u)z ,其中 u 为电站结点的序号, 0≤z≤10000 ,代表 pmax(u) 的值;每个测试数据的最后是 nc 个二元组, (u)z ,其中 u 为消费者结点的序号, 0≤z≤10000 ,代表 cmax(u) 的值。所有数据都是整数。除三元组 (u,v)z 和二元组 (u)z 中不含空格外,输入文件中其他位置允许出现空格。测试 数据一直到文件末尾。

 

思路

建立一个超级源点,发电站所提供的电量可以看做是从超级源点与发电站之间边的容量,同样建立一个超级汇点,消费者所消耗的电量可以看做是从它与超级汇点之间边的容量,对于三元组,u到v可以传输的电量便可以看做u到v之间一条容量为z的边。

最后求从超级源点到超级汇点之间的最大流即可。

 

AC代码

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<vector>
#include<queue>
using namespace std;

const int MAXN = 110;
const int MAXM = 110*110*2;
const int INF = 0x3f3f3f3f;

struct edge
{
    int v,f;    //边终点与当前流量
    int next;   //下一个兄弟位置
} edge[MAXM];

int n,np,nc,m;
int index;
int head[MAXN];
int level[MAXN];

void init()
{
    index =0;
    memset(head,-1,sizeof(head));
}

void addedge(int u,int v,int f) //同时往图中添加原边与反向边
{
    edge[index].v=v;
    edge[index].f=f;
    edge[index].next=head[u];
    head[u]=index++;
    edge[index].v=u;
    edge[index].f=0;
    edge[index].next=head[v];
    head[v]=index++;
}

int bfs(int s,int t)    //利用bfs搜索当前残量图中是否存在s到t的路径
{
    memset(level,0,sizeof(level));  //level为从源点到当前点的距离
    level[s]=1;
    queue<int>q;
    q.push(s);
    while(!q.empty())
    {
        int x=q.front();
        q.pop();
        if(x==t)return 1;
        for(int e=head[x]; e!=-1; e=edge[e].next)  //遍历所有与u临接的点
        {
            int v=edge[e].v,f=edge[e].f;
            if(!level[v]&&f)
            {
                level[v]=level[x]+1;    //广度优先搜索下一层距离源点比当前多1
                q.push(v);
            }
        }
    }
    return 0;
}

int dfs(int s,int maxf,int t)
{
    if(s==t)return maxf;    //找到一条路径,返回当前路径中的最小流量
    int ret=0;
    for(int e=head[s]; e!=-1; e=edge[e].next)   //遍历s所有邻接的点
    {
        int v=edge[e].v,f=edge[e].f;
        if(level[s]+1==level[v]&&f) //如果新点可达并且标号为允许弧
        {
            int minn=min(maxf-ret,f);
            f=dfs(v,minn,t);
            edge[e].f-=f;
            edge[e^1].f+=f;
            ret+=f;
            if(ret==maxf)return ret;
        }
    }
    return ret;
}

int dinic(int s,int t)
{
    int ans=0;
    while(bfs(s,t)) //先确定是否存在路径从s到t
        ans+=dfs(s,INF,t);  //
    return ans;
}

int main()
{
    while(~scanf("%d%d%d%d",&n,&np,&nc,&m)) //n作为源点,n+1作为汇点
    {
        init();
        int ai,bi,ci;
        for(int i=0; i<m; i++)  //ai到bi最大传输量为ci
        {
            while(getchar()!='(');
            scanf("%d%*c%d%*c%d",&ai,&bi,&ci);
            addedge(ai,bi,ci);
        }
        for(int i=0; i<np; i++) //ai最大发电量为bi
        {
            while(getchar()!='(');
            scanf("%d%*c%d",&ai,&bi);
            addedge(n,ai,bi);
        }
        for(int i=0; i<nc; i++) //ai最大消耗量为bi
        {
            while(getchar()!='(');
            scanf("%d%*c%d",&ai,&bi);
            addedge(ai,n+1,bi);
        }
        printf("%d\n",dinic(n,n+1));
    }
    return 0;
}