poj3436 ACM Computer Factory 拆點+網路流

ACM Computer Factory

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4674		Accepted: 1582		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,1Si,2...Si,PDi,1Di,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 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

    由於該題是節點流量限制，所以應該把每個點拆成兩個點，從而轉化成邊流量限制。

建圖：將每台機器看做一個節點，然後把每個節點拆成兩個（i和i+n），建邊i->i+n，邊流量限制為該機器的速度；對於每台機器i，若s[j]=0||s[j]=2,則建邊s->i,邊流量限制為INF，若d[j]=1，則建邊i+n->t,邊流量限制為INF；然後枚舉所有的機器對（i，j），若符合要求則建邊i+n->j，邊流量限制為INF。

    然後就可以求最大流了，最後輸出所有流量大於0的邊以及流量。

#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;}