poj 3436(网络流)

这题也算是一个简单的网络流吧， 但是题目的意思看了好久才看懂， 然后一开始我还想错了， 一拿到题目我就往二进制转换上想了，所以想了很久都没有想出来，关键是最后哪一步要求出每次转移的机器， 如果用一种状态用一个数表示的话就很难求，。

后面看了下别人的才恍然大悟， 直接以每台机器作为点，然后拆点，连边就行了， 一般的网络流的做法。

还有一个就是怎么求出从那台机器转移到那台机器， 可以在用网络流记录有流流过的边。 然后再选出一些符合要求的就行了.

 

ACM Computer Factory
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 4425   Accepted: 1489   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
 

 

  1 #include <stdio.h>
  2 #include <string.h>
  3 #include <string>
  4 #include <iostream>
  5 using namespace std;
  6 #define INF 0x3ffffff
  7 #define N 55
  8 struct node
  9 {
 10     int to,next,w,from;
 11     int sign;
 12 }edge[10*N*N];
 13 // 你个2货，完写错了
 14 
 15 struct node1
 16 {
 17     int u,v,w;
 18 }save[N*N];
 19 
 20 int p,n;
 21 int nn;
 22 int s,t;
 23 int cnt,pre[N];
 24 int saves[N][11],savet[N][11],savew[N];
 25 int lv[N],gap[N];
 26 
 27 void add_edge(int u,int v,int w)
 28 {
 29     edge[cnt].from=u;
 30     edge[cnt].sign=0;
 31     edge[cnt].to=v;
 32     edge[cnt].w=w;
 33     edge[cnt].next=pre[u];
 34     pre[u]=cnt++;
 35 }
 36 
 37 int sdfs(int k,int w)
 38 {
 39     if(k==t) return w;
 40     int f=0;
 41     int mi=nn-1;
 42     for(int p=pre[k];p!=-1;p=edge[p].next)
 43     {
 44         int v=edge[p].to;
 45         if(edge[p].w!=0)
 46         {
 47             if(lv[k]==lv[v]+1)
 48             {
 49                 int tmp=sdfs(v,min(w-f,edge[p].w));
 50                 if(tmp!=0)
 51                     edge[p].sign=1;
 52                 f+=tmp;
 53                 edge[p].w-=tmp;
 54                 edge[p^1].w+=tmp;
 55                 if(f==w||lv[s]==nn) return f;
 56             }
 57             if(lv[v]<mi) mi=lv[v];
 58         }
 59     }
 60     if(f==0)
 61     {
 62         gap[lv[k]]--;
 63         if(gap[lv[k]]==0)
 64         {
 65             lv[s]=nn;
 66             return f;
 67         }
 68         lv[k]=mi+1;
 69         gap[lv[k]]++;
 70     }
 71     return f;
 72 }
 73 
 74 int sap()
 75 {
 76     nn=t+1;
 77     int sum=0;
 78     memset(lv,0,sizeof(lv));
 79     memset(gap,0,sizeof(gap));
 80     gap[0]=nn;
 81     while( lv[s]<nn )
 82     {
 83         sum += sdfs(s,INF);
 84     }
 85     return sum;
 86 }
 87 
 88 /*void dfs(int k,int sum,int w)
 89 {
 90     if(k==p)
 91     {
 92         sum++;
 93         if(sum==1)
 94         {
 95             add_edge(s,sum,w);
 96             add_edge(sum,s,0);
 97         }
 98         add_edge(sum,nn,w);
 99         add_edge(nn,sum,0);
100         return ;
101     }
102     if(saves[k]==2)
103     {
104         dfs(k+1,sum*2+0,w);
105         dfs(k+1,sum*2+1,w);
106     }
107     else
108     {
109         dfs(k+1,sum*2+saves[k],w);
110     }
111 }*/
112 
113 int main()
114 {
115     memset(pre,-1,sizeof(pre));
116     cnt=0;
117     scanf("%d%d",&p,&n);
118     s=0;
119     t=2*n+1;
120     for(int i=1;i<=n;i++)
121     {
122         scanf("%d",&savew[i]);
123         for(int j=0;j<p;j++)
124             scanf("%d",&saves[i][j]);
125         for(int j=0;j<p;j++)
126             scanf("%d",&savet[i][j]);
127     }
128     for(int i=1;i<=n;i++) // 构图, 以后还是用n个一次的构图方法吧
129     {
130         add_edge(i,n+i,savew[i]);
131         add_edge(n+i,i,0);
132         int flag=0;
133         for(int j=0;j<p;j++)
134         {
135             if(saves[i][j]==2) continue;
136             if(saves[i][j]!=0)
137             {
138                 flag=1;  break;
139             }
140         }
141         if(flag==0)
142         {
143             add_edge(s,i,INF);
144             add_edge(i,s,0);
145         }
146         flag=0;
147         for(int j=0;j<p;j++)
148         {
149             if(savet[i][j]!=1)
150             {
151                 flag=1;
152                 break;
153             }
154         }
155         if( flag==0 )
156         {
157             add_edge(n+i,t,INF);
158             add_edge(t,n+i,0);
159         }
160         ///////////////////////////////
161         for(int j=1;j<=n;j++)
162         {
163             if(i==j) continue;
164             flag=0;
165             for(int k=0;k<p;k++)
166             {
167                 if( saves[j][k]== 2) continue;
168                 if(saves[j][k]!=savet[i][k])
169                 {
170                     flag=1;
171                     break;
172                 }
173             }
174             if(flag==0)
175             {
176                 add_edge(n+i,j,INF);
177                 add_edge(j,n+i,0);
178             }
179         }
180     }
181     printf("%d ",sap());
182     int tcnt=0;
183     for(int i=0;i<cnt;i+=2)
184     {
185         if( edge[i].sign == 1 )
186         {
187             int x,y;
188             x=edge[i].from;
189             y=edge[i].to;
190             if(y==t||x==0) continue;
191             if(x>n) x-=n;
192             if(y>n) y-=n;
193             if(x==y) continue;
194             save[tcnt].u=x;
195             save[tcnt].v=y;
196             save[tcnt].w=edge[i^1].w;
197             tcnt++;
198         }
199     }
200     printf("%d\n",tcnt);
201     for(int i=0;i<tcnt;i++)
202     {
203         printf("%d %d %d\n",save[i].u,save[i].v,save[i].w);
204     }
205     return 0;
206 }

 

posted @ 2013-02-06 23:25  chenhuan001  阅读(164)  评论(0)    收藏  举报