Poj3436 ACM computer factory split point + network stream

ACM computer factory

Time limit:1000 ms		Memory limit:65536 K
Total submissions:4674		Accepted:1582		Special Judge

Description

As you know, all the computers used for ACM contests must be identical, so the participates in 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
NVarious 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
PNumbers 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
PNumbers 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 your 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, ThenNDescriptions of the machines. The descriptionITh machine is represented as by 2
P+ 1 IntegersQi Si, 1Si, 2...Si,PDi, 1Di, 2...Di,P, Where
QiSpecifies 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, thenMDescriptions of the connections. Each connection between machines
AAndBMust be described by three positive numbersA B W, Where
WIs the number of computers delivered fromAToBPer hour.

If several solutions exist, output any of them.

Sample Input

Sample input 13 415  0 0 0  0 1 010  0 0 0  0 1 130  0 1 2  1 1 13   0 2 1  1 1 1Sample input 23 55   0 0 0  0 1 0100 0 1 0  1 0 13   0 1 0  1 1 01   1 0 1  1 1 0300 1 1 2  1 1 1Sample input 32 2100  0 0  1 0200  0 1  1 1

Sample output

Sample output 125 21 3 152 3 10Sample output 24 51 3 33 5 31 2 12 4 14 5 1Sample output 30 0

Because this is a node traffic limit, we should split each vertex into two vertices to convert them into edge traffic limits.

Build a graph: Think of each machine as a node, split each node into two (I and I + n), and create edge I-> I + N, edge traffic limit is the speed of the machine. For each machine I, if s [J] = 0 | s [J] = 2, the edge S-> I, the edge traffic limit is INF. If D [J] = 1, the edge I + N-> T is created, and the edge traffic limit is INF. Then, all the machine pairs (I, j). If yes, edge I + N-> J is created, and the edge traffic limit is INF.

Then we can find the maximum stream, and finally output all the edge and traffic with the traffic greater than 0.

#include<iostream>#include<cstring>#include<cstdio>using namespace std;const int MAXN=55;const int INF=(1<<29);struct{    int q;    int s[10],d[10];}machine[MAXN];struct Arc{    int c,f;};Arc flow[MAXN*2][MAXN*2];int level[MAXN*2];int p,n,s,t;int bfs(){    int queue[MAXN*2],front,rear;    front=rear=0;    memset(level,0,sizeof(level));    level[s]=1;    queue[rear++]=s;    while(front!=rear)    {        int v=queue[front++];        for(int i=0;i<=t;i++)        {            if(!level[i]&&flow[v][i].f<flow[v][i].c)            {                level[i]=level[v]+1;                queue[rear++]=i;            }        }    }    return level[t];}int dfs(int i,int f){    if(i==t)        return f;    int sum=0;    for(int j=0;f&&j<=t;j++)    {        if(level[j]==level[i]+1&&flow[i][j].f<flow[i][j].c)        {            int tmp=dfs(j,min(f,flow[i][j].c-flow[i][j].f));            sum+=tmp;            f-=tmp;            flow[i][j].f+=tmp;            flow[j][i].f-=tmp;        }    }    return sum;}int dinic(){    int maxflow=0;    while(bfs())        maxflow+=dfs(s,INF);    return maxflow;}int main(){    int i,j,k;    int maxflow,route;    while(~scanf("%d%d",&p,&n))    {        for(i=1;i<=n;i++)        {            scanf("%d",&machine[i].q);            for(j=0;j<p;j++)                scanf("%d",&machine[i].s[j]);            for(j=0;j<p;j++)                scanf("%d",&machine[i].d[j]);        }        s=0;        t=2*n+1;        for(i=0;i<=t;i++)            for(j=0;j<=t;j++)                flow[i][j].c=flow[i][j].f=0;        for(i=1;i<=n;i++)            flow[i][i+n].c=machine[i].q;        for(i=1;i<=n;i++)        {            int sum=0;            for(j=0;j<p;j++)            {                if(machine[i].s[j]==2)                    continue;                sum+=machine[i].s[j];            }            if(sum==0)                flow[s][i].c=INF;            sum=0;            for(j=0;j<p;j++)                sum+=machine[i].d[j];            if(sum==p)                flow[i+n][t].c=INF;        }        for(i=1;i<=n;i++)            for(j=1;j<=n;j++)            {                if(i==j)                    continue;                k=0;                while(k<p&&(machine[i].d[k]==machine[j].s[k]||machine[j].s[k]==2))                    k++;                if(k==p)                    flow[i+n][j].c=INF;            }        maxflow=dinic();        route=0;        for(i=n+1;i<=t;i++)            for(j=1;j<=n;j++)                if(i-n!=j&&flow[i][j].f)                    route++;        printf("%d %d\n",maxflow,route);        for(i=n+1;i<=t;i++)            for(j=1;j<=n;j++)                if(j!=i-n&&flow[i][j].f)                    printf("%d %d %d\n",i-n,j,flow[i][j].f);    }    return 0;}