POJ3436：ACM Computer Factory（最大流）

ACM Computer Factory

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9963		Accepted: 3738		Special Judge

题目链接：http://poj.org/problem?id=3436

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 Q_i S_i,1 Si_,2...S_i,P D_i,1 D_i,2...D_i,P, where Q_i specifies performance, S_i,j — input specification for part j, D_i,k — output specification for part k.

Constraints

1 ≤ P ≤ 10, 1 ≤ N ≤ 50, 1 ≤ Q_i ≤ 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

Hint

Bold texts appearing in the sample sections are informative and do not form part of the actual data.

 

题意：

这个题意难度有点大呀TnT。

简单说就是给出n个机器，每个机器都有一个pi代表生产效率，然后后面有2*p个数，前p个代表输入规范，后p个代表输出规范。这里输入输出规范的意思就是这台机器可以把输入规范转化为输出规范。

输入规范里面0就是这里没有零件，1就是这里必须有零件，2就是这里可有可不有；输出规范里面就只有0，1，意义同上。

现在问最多可以生产出多少台电脑（假设机器的配合不消耗时间= =），只有输出规范全为1的机器可以生产电脑。

 

题解：

这题用最大流来做。首先建立一个超级源点和超级汇点，超级源点连上输入规范全为0，或有0也有2的机器，因为这些机器可以“无中生有”，边权为无穷大。

然后所有输出规范为1的机器就连向超级汇点，毕竟此时可以生成电脑，边权也为无穷大。

然后建立可以相互可达的机器之间的边，这里由于每个点有个生产效率的权值，所以我们考虑把点拆开为一条权值为其生产效率的有向边，拆成的两个点分别代表入读点和出度点。这样可以限定一条生产线上的生产效率。

然后直接跑最大流就好了~

最后统计结果的时候就随便统计一下就好了...毕竟special judge。如果一个出度点到一个入读点的边上面有流量，就代表了两个点之间有合作关系，就输出这两个点。

代码如下：

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <vector>
#include <queue>
#define INF 99999999
using namespace std;

const int N = ;
int P,n,tot;
int p[N],m[N][N],head[N],cur[N],d[N];

struct Edge{
    int u,v,c,flow,next;
}e[N<<];
void adde(int u,int v,int w,int f){
    e[tot].v=v;e[tot].u=u;e[tot].c=w;e[tot].flow=f;
    e[tot].next=head[u];head[u]=tot++;
}
bool bfs(int s,int t){
    for(int i=;i<=*n+;i++) d[i]=;d[s]=;
    queue <int > q;q.push(s);
    while(!q.empty()){
        int u=q.front();q.pop();
        for(int i=head[u];i!=-;i=e[i].next){
            int v=e[i].v;
            if(!d[v] && e[i].c>e[i].flow){
                d[v]=d[u]+;
                q.push(v);
            }
        }
    }
    return d[t]!=;
}
int dfs(int s,int a){
    if(s==*n+ || a==) return a;
    int flow = ;
    for(int &i=cur[s];i!=-;i=e[i].next){
        int v=e[i].v,f;
        if(d[v]!=d[s]+) continue ;
        f=dfs(v,min(a,e[i].c-e[i].flow));
        if(f){
            e[i].flow+=f;
            e[i^].flow-=f;
            a-=f;
            flow+=f;
            if(a==) break;
        }
    }
    if(!flow) d[s]=-;
    return flow;
}
int main(){
    scanf("%d%d",&P,&n);
    memset(head,-,sizeof(head));
    for(int i=;i<=n;i++){
        scanf("%d",&p[i]);
        for(int j=;j<=*P;j++) scanf("%d",&m[i][j]);
    }

    for(int i=;i<=n;i++){
        adde(i,i+n,p[i],);
        adde(i+n,i,,);
        int flag1=,flag2=;
        for(int j=;j<=P;j++){
            if(m[i][j]==) flag1=;
            if(m[i][j+P]!=) flag2=;
        }
        if(flag1) adde(,i,INF,),adde(i,,,);
        if(flag2) adde(i+n,*n+,INF,),adde(*n+,i+n,,);
    }
    for(int i=;i<=n;i++){
        for(int j=;j<=n;j++){
            if(i==j) continue ;
            bool ok = true ;
            for(int k=P+;k<=P*;k++){
                int now = k-P;
                if(m[j][now]==) continue ;
                if(m[i][k]!=m[j][now]) ok=false;
            }
            if(ok){
                adde(i+n,j,INF,);
                adde(j,i+n,,);
            }
        }
    }

    int max_flow = ;
    while(bfs(,*n+)){
        for(int i=;i<=*n+;i++) cur[i]=head[i];
        max_flow+=dfs(,INF);
    }
    printf("%d ",max_flow);
    int tot=;
    vector <pair<int,int> > ans[N];
    for(int i=+n;i<=*n;i++){
        for(int j=head[i];j!=-;j=e[j].next){
            int v=e[j].v;
            if(v!=*n+ && v!= && e[j].flow && v!=i-n) ans[i-n].push_back(make_pair(v,e[j].flow)),tot++;
        }
    }
    printf("%d\n",tot);
    for(int i=;i<=n;i++)
        for(int j=;j<ans[i].size();j++){
            printf("%d %d %d\n",i,ans[i][j].first,ans[i][j].second);
        }
    return ;
}