POJ3436------ACM Computer Factory
ACM Computer Factory
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.
题意：有N台机器生产电脑，每个电脑有p个部件，，同时每个机器都有一个效率。不同机器能够安装的部件不一定一样，面对不同的机器，我们需要让两个机器对接起来。
题解： 这个题需要通过拆点去做，机器与机器是之间的传输是无穷大的但是机器内部并不是，下面是我画的第一个样例的图示
我用的是EK算法。代码不够精炼，还望海涵。
#include<stdio.h>
#include<string.h>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=100;
int in[55][20],out[50][20];
int cap[MAXN][MAXN];
int pre[MAXN];
bool vis[MAXN];
int p,n;
int N=n*2;
bool match(int *a,int *b)
{
for(int i=1;i<=p;i++)
{
if(a[i]==b[i]) continue;
if(b[i]==2) continue;
return false;
}
return true;
}
bool matchs(int *a)
{
for(int i=1;i<=p;i++)
{
if(a[i]==1)
return false;
}
return true;
}
bool matcht(int *b)
{
for(int i=1;i<=p;i++)
{
if(b[i]==0)
return false;
}
return true;
}
bool bfs(int s,int t)
{
memset(pre, -1, sizeof(pre));
queue<int> que;
memset(vis,0 , sizeof(vis));
vis[s]=true;
pre[s]=s;
que.push(s);
int p;
while(!que.empty())
{
p=que.front();
que.pop();
for(int i=0;i<=N;i++)
{
if(!vis[i]&&cap[p][i]>0)
{
vis[i]=true;
pre[i]=p;
if(i==t) return true;
que.push(i);
}
}
}
return false;
}
int EK(int s,int t)
{
int ans=0,d;
while (bfs(s,t))
{
d=INF;
for(int i=t;i!=s;i=pre[i])
d=min(d,cap[pre[i]][i]);
for(int i=t;i!=s;i=pre[i])
{
cap[pre[i]][i]-=d;
cap[i][pre[i]]+=d;
}
ans+=d;
}
return ans;
}
int main()
{
while(scanf("%d%d",&p,&n )!=EOF)
{
memset(cap,0 , sizeof(cap));
int c;
for(int i=1;i<=n;i++)
{
scanf("%d",&c);
cap[2*i-1][2*i]=c;
for(int j=1;j<=p;j++)
scanf("%d",&in[i][j]);
for(int j=1;j<=p;j++)
scanf("%d",&out[i][j]);
}
N=2*n+1;
for(int i=1;i<=n;i++)
{
if(matchs(in[i]))
{
cap[0][2*i-1]=INF;
// printf("%d %d---\n",0,2*i-1);
}
if(matcht(out[i]))
{
cap[2*i][N]=INF;
// printf("%d %d++++\n",2*i,N);
}
for(int j=1;j<=n;j++)
{
if(match(out[i],in[j]))
{
cap[2*i][2*j-1]=INF;
// printf("%d %d##\n",2*i,2*j-1);
}
}
}
printf("%d ",EK(0,N));
int ans=0;
int x[MAXN],y[MAXN],z[MAXN];
for (int i = 1; i <n; ++i)
{
for(int j=1;j<=n;j++)
{
if(match(out[i],in[j]))
{
if(cap[2*i][2*j-1]!=INF)
{
ans++;
x[ans-1]=i;y[ans-1]=j;z[ans-1]=INF-cap[2*i][2*j-1];
if(z[ans-1]<0)//此处是考虑会把回去的边计算上，我们需要把他去掉。
ans--;
}
}
}
}
printf("%d\n",ans);
for(int i=0;i<ans;i++)
printf("%d %d %d\n",x[i],y[i],z[i]);
}
return 0;
}
