POJ 3436 ACM Computer Factory(網路最大流)

標籤：algorithm   圖論   網路流   

http://poj.org/problem?id=3436

ACM Computer Factory Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5286   Accepted: 1813   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 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 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：

流水線上有N台機器裝電腦，電腦有P個組件，每台機器有三個參數，產量，輸入規格，輸出規格;輸入規格中0表示改組件不能有，1表示必須有，2無所謂;輸出規格中0表示改組件沒有，1表示有。問如何安排流水線（如何建邊）使產量最高。

分析：

這是道網路最大流

首先拆點，因為點上有容量限制，在拆出的兩點間連一條邊，容量為產量;

如果輸入規格中沒有1,則與超級源點相連，容量為無窮大;

如果輸出規格中沒有0,則與超級匯點相連，容量為無窮大;

判斷和其他機器的輸入輸出是否匹配，能匹配上則建邊，容量為無窮大;

在該圖上跑一遍最大流，容量大於零的邊集即為方案。

#include<cstdio>#include<iostream>#include<cstdlib>#include<algorithm>#include<ctime>#include<cctype>#include<cmath>#include<string>#include<cstring>#include<stack>#include<queue>#include<list>#include<vector>#include<map>#include<set>#define sqr(x) ((x)*(x))#define LL long long#define itn int#define INF 0x3f3f3f3f#define PI 3.1415926535897932384626#define eps 1e-10#define maxm 210007#define maxn 107using namespace std;int fir[maxn];int u[maxm],v[maxm],cap[maxm],flow[maxm],nex[maxm];int e_max;int lv[maxn],q[maxn],iter[maxn];void add_edge(int _u,int _v,int _w){    int e;    e=e_max++;    u[e]=_u;v[e]=_v;cap[e]=_w;nex[e]=fir[u[e]];fir[u[e]]=e;    e=e_max++;    u[e]=_v;v[e]=_u;cap[e]=0;nex[e]=fir[u[e]];fir[u[e]]=e;}void dinic_bfs(int s){    int f,r;    memset(lv,-1,sizeof lv);    lv[s]=0;    q[f=r=0]=s;    while (f<=r)    {        int x=q[f++];        for (int e=fir[x];~e;e=nex[e])        {            if(cap[e]>flow[e] && lv[v[e]]<0)            {                lv[v[e]]=lv[u[e]]+1;                q[++r]=v[e];            }        }    }}int dinic_dfs(int _u,int t,int _f){    if (_u==t)  return _f;    for (int &e=iter[_u];~e;e=nex[e])    {        if (cap[e]>flow[e] && lv[u[e]]<lv[v[e]])        {            int _d=dinic_dfs(v[e],t,min(_f,cap[e]-flow[e]));            if (_d>0)            {                flow[e]+=_d;                flow[e^1]-=_d;                return _d;            }        }    }    return 0;}int max_flow(int s,int t){    memset(flow,0,sizeof flow);    int total_flow=0;    for (;;)    {        dinic_bfs(s);        if (lv[t]<0)    return total_flow;        memcpy(iter,fir,sizeof fir);        int _f;        while ((_f=dinic_dfs(s,t,INF))>0)            total_flow+=_f;    }    return total_flow;}struct node{    int in[10],out[10];    int in1,out0;}a[55];int main(){    #ifndef ONLINE_JUDGE        freopen("/home/fcbruce/文檔/code/t","r",stdin);    #endif // ONLINE_JUDGE    int p,n,s,t,_w;    e_max=0;    memset(fir,-1,sizeof fir);    scanf("%d %d",&p,&n);    s=0;t=n+n+1;    for (int i=1;i<=n;i++)    {        scanf("%d",&_w);        add_edge(i,i+n,_w);        a[i].in1=0;        for (int j=0;j<p;j++)        {            scanf("%d",&a[i].in[j]);            if (a[i].in[j]==1)  a[i].in1++;        }        a[i].out0=0;        for (int j=0;j<p;j++)        {            scanf("%d",&a[i].out[j]);            if (a[i].out[j]==0)  a[i].out0++;        }        if (a[i].in1==0)    add_edge(s,i,INF);        if (a[i].out0==0)    add_edge(i+n,t,INF);        for (int j=1;j<i;j++)        {            if (a[j].in1!=0 && a[i].out0!=0)            {                bool flag=true;                for (int k=0;flag && k<p;k++)                {                    if (a[i].out[k]+a[j].in[k]==1)                        flag=false;                }                if (flag)                    add_edge(i+n,j,INF);            }            if (a[i].in1!=0 && a[j].out0!=0)            {                bool flag=true;                for (int k=0;flag && k<p;k++)                {                    if (a[j].out[k]+a[i].in[k]==1)                        flag=false;                }                if (flag)                    add_edge(j+n,i,INF);            }        }    }    printf("%d ",max_flow(s,t));    int m=0;    for (int e=0;e<e_max;e++)    {        if (flow[e]<=0 || u[e]==s || v[e]==t || v[e]-u[e]==n) continue;        u[m]=u[e]>n?u[e]-n:u[e];        v[m]=v[e];        flow[m]=flow[e];        m++;    }    printf("%d\n",m);    for (int e=0;e<m;e++)        printf("%d %d %d\n",u[e],v[e],flow[e]);    return 0;}