ACM Computer Factory - poj 3436 （最大流）

 

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5949		Accepted: 2053		Special Judge

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：每台电脑有P部分，可以通过不同的机器来进行加工。 有N台机器，每台机器用2 P +1 个整数来描述：Qi Si,1 Si,2 ... Si,p Di,1 Di,2. .. Di,p，其中Qi 指定了机器的性能，表示每小时加工的电脑数量。 Si,j 为第j 部分的输入规格，0表示该部分不能被加工过，1表示该部分必须被加工过，2表示都可以。 Di,k 为第k 部分的输出规格。 0表示经过该机器不加工，1表示该机器加工该部分。 1≤P≤10，1≤N≤50，1≤Qi≤10000。

注意：本题Sample I/O这段英文不用输入输出

Sample input:

P  N （N台机器，每台机器有P部分）

接着输入N行，其实每行都是一个结点的信息

每一行的格式为 一个Q  P个S  P个D

其中Q为当前结点的容量，S都是当前结点的输入规格，D都是输出规格

Sample output:

第一行的两个数字分别表示：最大流的值，流量发生变化的边数M（和s还有t关联的边不在其内，那些不属于原有的边，是附加边）

接下来有M行，每一行都有三个数字，A B W

A B为流量发生变化的边的端点，W为流量的变化值（每条边初始流量为0，最终流量就是找到最大流时的流量）

若图不连通，则输出0 0

 #include <iostream>
 #include<stdio.h>
 #include<queue>
 #include<string.h>
 using namespace std;
 int map[][],mapbk[][];
 int input[][];
 int path[];
 int flow[];
 int change[][];
 int start,end;
 int p,n;
 queue<int> q;
 int BFS(){
     memset(path,-,sizeof(path));
     while(!q.empty()) q.pop();
     q.push(start);
     flow[start]=;
     path[start]=;
     while(!q.empty()){
         int v=q.front();
         if(v==end)
             break;
         q.pop();
         for(int i=;i<=end;i++){
             if(path[i]==- && map[v][i]!= && start!=i){
                 flow[i]=flow[v]<map[v][i]?flow[v]:map[v][i];
                 path[i]=v;
                 q.push(i);
             }
         }

     }
     if(path[end]==-) return -;
     else
         return flow[end];
 }
 int Edmonds_Karp(){
     int step,max_flow=,now,pre;
         while(){
             step=BFS();
             if(step==-)
                 break;
             max_flow+=step;
             now=end;
             while(now!=start){
                 pre=path[now];
                 map[pre][now]-=step;
                 map[now][pre]+=step;
                 now=pre;
             }

         }
         return max_flow;
 }
 int main() {
     while(cin>>p>>n){
         memset(map,,sizeof(map));
         memset(input,,sizeof(input));
         //gets(str);
         for(int i=;i<=n;i++){
             for(int j=;j<*p+;j++){
                 int t;
                 cin>>t;
                 input[i][j]=t;
             }
         }
                     //getchar();
                 //    gets(str);
         for(int i=;i<=n;i++){
             int flag=;
             for(int j=;j<p+;j++){
                 if(input[i][j]==)
                     flag=;
             }
             if(flag!=){
                 map[][i]=input[i][];
             }
             flag=;
             for(int j=p+;j<*p+;j++){
                 if(input[i][j]==)
                     flag=;
             }
             if(flag!=)
                 map[i][n+]=input[i][];
             flag=;
             for(int j=;j<=n;j++){
                 if(i!=j){
                     for(int k=;k<p+;k++){
                         if(i!=j&&input[i][p+k]+input[j][k]==)
                             flag=;
                         }
                         if(flag!=)
                         map[i][j]=input[i][]<input[j][]?input[i][]:input[j][];
                         flag=;
                         }
                 }

         }
         start=;
         end=n+;
         memcpy(mapbk,map,sizeof(map));
         int result=Edmonds_Karp();

         int count=;
         for(int i=;i<n+;i++){
             for(int j=;j<n+;j++){
                 if(map[i][j]<mapbk[i][j]){
                     change[count][]=i;
                     change[count][]=j;
                     change[count][]=mapbk[i][j]-map[i][j];
                     count++;
                 }
             }
         }
         cout<<result<<' '<<count<<endl;
         for(int i=;i<count;i++){
             cout<<change[i][]<<' '<<change[i][]<<' '<<change[i][]<<endl;
         }
     }
     return ;
 }