POJ 3436 ACM Computer Factory (网络流,最大流)

POJ 3436 ACM Computer Factory (网络流,最大流)

Description

As you know, all the computers used for ACM contests must be identical, so the participants compete on equal terms. That is why all these computers are historically produced at the same factory.

Every ACM computer consists of P parts. When all these parts are present, the computer is ready and can be shipped to one of the numerous ACM contests.

Computer manufacturing is fully automated by using N various machines. Each machine removes some parts from a half-finished computer and adds some new parts (removing of parts is sometimes necessary as the parts cannot be added to a computer in arbitrary order). Each machine is described by its performance (measured in computers per hour), input and output specification.

Input specification describes which parts must be present in a half-finished computer for the machine to be able to operate on it. The specification is a set of P numbers 0, 1 or 2 (one number for each part), where 0 means that corresponding part must not be present, 1 — the part is required, 2 — presence of the part doesn't matter.

Output specification describes the result of the operation, and is a set of P numbers 0 or 1, where 0 means that the part is absent, 1 — the part is present.

The machines are connected by very fast production lines so that delivery time is negligibly small compared to production time.

After many years of operation the overall performance of the ACM Computer Factory became insufficient for satisfying the growing contest needs. That is why ACM directorate decided to upgrade the factory.

As different machines were installed in different time periods, they were often not optimally connected to the existing factory machines. It was noted that the easiest way to upgrade the factory is to rearrange production lines. ACM directorate decided to entrust you with solving this problem.

Input

Input file contains integers P N, then N descriptions of the machines. The description of ith machine is represented as by 2 P + 1 integers Qi Si,1 Si,2...Si,P Di,1 Di,2...Di,P, where Qi specifies performance, Si,j — input specification for part j, Di,k — output specification for part k.

Constraints

1 ≤ P ≤ 10, 1 ≤ N ≤ 50, 1 ≤ Qi ≤ 10000

Output

Output the maximum possible overall performance, then M — number of connections that must be made, then M descriptions of the connections. Each connection between machines A and B must be described by three positive numbers A B W, where W is the number of computers delivered from A to B per hour.

If several solutions exist, output any of them.

Sample Input

Sample input 1
3 4
15 0 0 0 0 1 0
10 0 0 0 0 1 1
30 0 1 2 1 1 1
3 0 2 1 1 1 1
Sample input 2
3 5
5 0 0 0 0 1 0
100 0 1 0 1 0 1
3 0 1 0 1 1 0
1 1 0 1 1 1 0
300 1 1 2 1 1 1
Sample input 3
2 2
100 0 0 1 0
200 0 1 1 1

Sample Output

Sample output 1
25 2
1 3 15
2 3 10
Sample output 2
4 5
1 3 3
3 5 3
1 2 1
2 4 1
4 5 1
Sample output 3
0 0

Http

POJ:https://vjudge.net/problem/POJ-3436

Source

网络流,最大流

题目大意

工厂里有若干机器人负责生产电脑,每个机器人负责加工一部分(这个加工可以是加上一个零件,也可以是减少一个零件),同时,每个机器人对进入其加工的电脑也有要求,对于每一个零件,1代表必须存在才能加工,0代表必须不存在才能加工,而2代表无所谓。每个机器人加工的效率是不一样的。现在给出所有机器人对进入的电脑的要求和出口的形式,以及加工效率,求最大的生产量。

解决思路

题目不是很好理解,需要多读几遍。
我们可以用网络流来解决这道题。
对于每一个机器人i,我们把它拆成两个点(笔者用i和i+n表示,其中n是所有机器人的数量),i表示入口,i+n表示出口并连一条边i->(i+n) ,流量就是机器人的工作效率。
我们再枚举每一对机器人i和j,若i的出口产品能被j接受,则在(i+n)->j之间连一条无穷大的边,当然也可以把边权赋为i与j中效率小的那个(因为再怎么也不会超过它)
然后我们再来考虑源点与汇点,笔者这里用0表示源点,n*2+1表示汇点。我们再枚举每一个机器人,如果某一个机器人的入口全部都是0,则从源点向它连一条无穷大的边;如果某一个机器人的出口全是1,则从它向汇点连一条无穷大的边。
这样,我们在这张图上跑一边最大流即可。
最后我们来考虑如何输出选择的连接方案。
我们可以在建完图后把原图备份一下,在跑完最大流后,将跑完后的图与原图对比,如果发现有流的减少,则说明走了这条边,统计一下即可。
虽然说笔者是使用EK算法实现最大流的,但更推荐时间效率更高的Dinic算法,具体请移步我的这篇文章

代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;

const int maxN=200;
const int maxP=30;
const int inf=2147483647;

class Edge//输出时要用
{
public:
    int u,v,w;
};

int n,m,P;
int G[maxN][maxN];//图
int G2[maxN][maxN];//备份的图,用来对比输出
int W[maxN];//每个机器人的效率
int In[maxN][maxP];//机器人对入口的要求
int Out[maxN][maxP];//机器人的出口规格
int Path[maxN];//EK算法中存增广路径
int Flow[maxN];//流
Edge Outp[maxN*maxN];//用于输出

bool bfs();//bfs增广

int main()
{
    memset(G,0,sizeof(G));
    scanf("%d%d",&P,&n);
    for (int i=1;i<=n;i++)//输入
    {
        scanf("%d",&W[i]);
        for (int j=1;j<=P;j++)
            scanf("%d",&In[i][j]);
        for (int j=1;j<=P;j++)
            scanf("%d",&Out[i][j]);
    }
    for (int i=1;i<=n;i++)//给每个点的入点到出点连一条边
        G[i][i+n]=W[i];
    for (int i=1;i<=n;i++)//给每对出口与入口相对应的点连一条边
        for (int j=1;j<=n;j++)
        {
            bool ok=1;
            for (int k=1;k<=P;k++)
                if (((Out[i][k]==1)&&(In[j][k]==0))||((Out[i][k]==0)&&(In[j][k]==1)))
                {
                    ok=0;
                    break;
                }
            if (ok)
                G[i+n][j]=min(W[i],W[j]);
        }
    for (int i=1;i<=n;i++)//判断能否和汇点或源点连接
    {
        bool ok=1;
        for (int j=1;j<=P;j++)
            if (In[i][j]==1)
            {
                ok=0;
                break;
            }
        if (ok)
            G[0][i]=inf;
        ok=1;
        for (int j=1;j<=P;j++)
            if (Out[i][j]==0)
            {
                ok=0;
                break;
            }
        if (ok)
            G[i+n][n*2+1]=inf;
    }
    memcpy(G2,G,sizeof(G2));//把G存起来,备份一边,方便后面输出答案
    int Ans=0;
    while (bfs())//笔者这里用EK算法实现最大流
    {
        int di=Flow[n*2+1];
        int now=2*n+1;
        int last=Path[now];
        while (now!=0)//更新残量网络
        {
            G[last][now]-=di;
            G[now][last]+=di;
            now=last;
            last=Path[now];
        }
        Ans+=di;
    }
    printf("%d ",Ans);//输出答案,统计路径
    int top=0;
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++)
            if (G2[i+n][j]>G[i+n][j])
            {
                top++;
                Outp[top].u=i;
                Outp[top].v=j;
                Outp[top].w=G2[i+n][j]-G[i+n][j];
            }
    printf("%d\n",top);//输出路径
    for (int i=1;i<=top;i++)
        printf("%d %d %d\n",Outp[i].u,Outp[i].v,Outp[i].w);
    return 0;
}

bool bfs()
{
    memset(Path,-1,sizeof(Path));
    memset(Flow,0,sizeof(Flow));
    queue<int> Q;
    while (!Q.empty())
        Q.pop();
    Q.push(0);
    Flow[0]=inf;
    do
    {
        int u=Q.front();
        Q.pop();
        for (int i=0;i<=2*n+1;i++)
        {
            if ((Path[i]==-1)&&(G[u][i]>0))
            {
                Path[i]=u;
                Q.push(i);
                Flow[i]=min(Flow[u],G[u][i]);
            }
        }
    }
    while (!Q.empty());
    if (Flow[2*n+1]>0)
        return 1;
    return 0;
}
posted @ 2017-07-29 14:00 SYCstudio 阅读(...) 评论(...) 编辑 收藏