ACM Computer Factory

ACM Computer Factory
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 6104 Accepted: 2113 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 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

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

Source
Northeastern Europe 2005, Far-Eastern Subregion
題意很差懂啊,看了許多的博客,才漸漸的理解題意了,但大多數的博客都須要拆點,但總感受不須要,就寫了一個簡單的Dinic,就AC,是否是數據水啊.

  1 #include <map>
  2 #include <list>
  3 #include <cmath>
  4 #include <queue>
  5 #include <stack>
  6 #include <string>
  7 #include <cstdio>
  8 #include <climits>
  9 #include <cstring>
 10 #include <cstdlib>
 11 #include <iostream>
 12 #include <algorithm>
 13 using namespace std;
 14 #define LL long long
 15 #define PI acos(-1.0)
 16 #define MMM 0x3f3f3f3f
 17 #define RR freopen("input.txt","r",stdin)
 18 #define WW freopen("output.txt","w",stdout)
 19 
 20 const int INF = 0x3f3f3f3f;
 21 
 22 struct node
 23 {
 24     int peed;
 25     int in[15];
 26     int out[15];
 27 } Point[55];
 28 int n,m;
 29 int s,t;
 30 int Map[55][55];
 31 int Flow[55][55];
 32 bool vis[55];
 33 bool sign[55][55];
 34 bool BFS()
 35 {
 36     memset(vis,false,sizeof(vis));
 37     memset(sign,false,sizeof(sign));
 38     queue<int>Q;
 39     Q.push(s);
 40     vis[s]=true;
 41     while(!Q.empty())
 42     {
 43         int u=Q.front();
 44         Q.pop();
 45         for(int i=0;i<=t;i++)
 46         {
 47             if(!vis[i]&&Map[u][i])
 48             {
 49                 sign[u][i]=true;
 50                 vis[i]=true;
 51                 Q.push(i);
 52             }
 53         }
 54     }
 55     return vis[t];
 56 }
 57 int DFS(int star,int num)
 58 {
 59     if(star==t||num==0)
 60     {
 61         return num;
 62     }
 63     int s=0;
 64     int ant;
 65     for(int i=0;i<=t;i++)
 66     {
 67         if(sign[star][i]&&(ant=DFS(i,min(Map[star][i],num)))>0)
 68         {
 69             Map[star][i]-=ant;
 70             Map[i][star]+=ant;
 71             Flow[star][i]+=ant;
 72             Flow[i][star]-=ant;
 73             num-=ant;
 74             s+=ant;
 75             if(num==0)
 76             {
 77                 break;
 78             }
 79         }
 80     }
 81     return s;
 82 }
 83 int Dinic()
 84 {
 85     memset(Flow,0,sizeof(Flow));
 86     int sum=0;
 87     while(BFS())
 88     {
 89         sum+=DFS(0,INF);
 90     }
 91     return sum;
 92 }
 93 int main()
 94 {
 95     while(~scanf("%d %d",&m,&n))
 96     {
 97         s=0;
 98         t=n+1;
 99         memset(Map,0,sizeof(Map));
100         for(int i=1; i<=n; i++)
101         {
102             scanf("%d",&Point[i].peed);
103             for(int j=1; j<=m; j++)
104             {
105                 scanf("%d",&Point[i].in[j]);
106             }
107             for(int j=1; j<=m; j++)
108             {
109                 scanf("%d",&Point[i].out[j]);
110             }
111         }
112         bool flag;
113         for(int i=1;i<=n;i++)
114         {
115             flag=false;
116             for(int j=1;j<=m;j++)
117             {
118                 if(Point[i].in[j]==1)
119                 {
120                     flag=true;
121                     break;
122                 }
123             }
124             if(!flag)
125             {
126                 Map[s][i]=Point[i].peed;
127             }
128             flag=false;
129             for(int j=1;j<=m;j++)
130             {
131                 if(Point[i].out[j]!=1)
132                 {
133                     flag=true;
134                     break;
135                 }
136             }
137             if(!flag)
138             {
139                 Map[i][t]=Point[i].peed;
140             }
141         }
142         for(int i=1;i<=n;i++)
143         {
144             for(int j=1;j<=n;j++)
145             {
146                 if(i!=j)
147                 {
148                     flag=false;
149                     for(int k=1;k<=m;k++)
150                     {
151                         if(Point[i].out[k]!=Point[j].in[k]&&Point[j].in[k]!=2)
152                         {
153                             flag=true;
154                             break;
155                         }
156                     }
157                     if(!flag)
158                     {
159                         Map[i][j]=min(Point[i].peed,Point[j].peed);
160                     }
161                 }
162             }
163         }
164         int sum=Dinic();
165         int num=0;
166         for(int i=1;i<=n;i++)
167         {
168             for(int j=1;j<=n;j++)
169             {
170                 if(Flow[i][j]>0)
171                 {
172                     num++;
173                 }
174             }
175         }
176         cout<<sum<<" "<<num<<endl;
177         for(int i =1;i<=n;i++)
178         {
179             for(int j=1;j<=n;j++)
180             {
181                 if(Flow[i][j]>0)
182                 {
183                     cout<<i<<" "<<j<<" "<<Flow[i][j]<<endl;
184                 }
185             }
186         }
187     }
188     return 0;
189 }

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