POJ 3436 ACM Computer Factory (最大流)

Posted 2020-07-30 季末Despaire

ACM Computer Factory

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6773		Accepted: 2379		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 Q_i S_i,1 Si_,2...S_i,P D_i,1 D_i,2...D_i,P, where Q_i specifies performance, S_i,j — input specification for part j, D_i,k — output specification for part k.

Constraints

1 ≤ P ≤ 10, 1 ≤ N ≤ 50, 1 ≤ Q_i ≤ 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

分析
题意就是一个电脑有P个部件，有N台机器，每个机器只能加工指定部件存在的半成品电脑，然后会加上或者减去电脑的一些部件，每台机器有一个performance，最后就是求最大的performance的和

  其中输入规格有三种情况：0,1,2

  0：该部分不能存在

  1：该部分必须保留

  2：该部分可有可无

  输出规格有2种情况：0,1

  0：该部分不存在

  1：该部分存在

       就是求最大流，关键在于建图，首先每个机器有一个performance，所以将每个机器拆成两个点，中间用边权为w的有向线段连接；

  输入规格全是0的和源点相连，全是1的和汇点相连。两台机器i和j，如果i的输出符合j的输入，就将他们连起来，取流量较小的一个

  然后跑最大流就行了。

  输出路径的话就先保存没跑最大流之前的图，和之后的比较，如果小，就说明这两个点是路径上的点，存起来输出就行了。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<queue>
using namespace std;
//****************************************************
//最大流模板Edmonds_Karp算法
//初始化：G[][],st,ed
//******************************************************
const int MAXN = 200+10;
const int INF = 0x3fffffff;
int G[MAXN][MAXN];//存边的容量，没有边的初始化为0
int path[MAXN],flow[MAXN],st,ed;
int n;//点的个数，编号0~n，n包括了源点和汇点

queue<int>q;
int bfs()
{
    int i,t;
    while(!q.empty()) q.pop();//清空队列
    memset(path,-1,sizeof(path));//每次搜索前都把路径初始化成-1
    path[st]=0;
    flow[st]=INF;//源点可以有无穷的流流进
    q.push(st);
    while(!q.empty()){
        t=q.front();
        q.pop();
        if(t==ed) break;
        for(i=0;i<=n;i++){
            if(i!=st&&path[i]==-1&&G[t][i]){
                flow[i]=flow[t]<G[t][i]?flow[t]:G[t][i];
                q.push(i);
                path[i]=t;
            }
        }
    }
    if(path[ed]==-1) return -1;//即找不到汇点上去了。找不到增广路径了
    return flow[ed];
}

int Edmonds_Karp()
{
    int max_flow=0;
    int step,now,pre;
    while((step=bfs())!=-1){
        max_flow+=step;
        now=ed;
        while(now!=st){
            pre=path[now];
            G[pre][now]-=step;
            G[now][pre]+=step;
            now=pre;
        }
    }
    return max_flow;
}

int in[MAXN][20];//输入信息
int backup[MAXN][MAXN];//备份图
int Line[MAXN][4];

int main()
{
    int P,N;
    while(scanf("%d%d",&P,&N)!=EOF){
        memset(G,0,sizeof(G));
        for(int i=1;i<=N;i++)
            for(int j=0;j<2*P+1;j++)
                scanf("%d",&in[i][j]);
        for(int i=1;i<=N;i++)//拆点
            G[2*i-1][2*i]=in[i][0];
        n=2*N+1;
        st=0,ed=n;
        for(int i=1;i<=N;i++){
            bool flag_s=true;
            bool flag_t=true;
            for(int j=1;j<=P;j++){
                if(in[i][j]==1) flag_s=false;
                if(in[i][j+P]==0) flag_t=false;
            }
            if(flag_s) G[st][2*i-1]=INF;
            if(flag_t) G[2*i][ed]=INF;
            for(int j=1;j<=N;j++){
                if(j==i) continue;
                bool flag=true;
                for(int k=1;k<=P;k++)
                    if((in[i][k+P]==0&&in[j][k]==1)||(in[i][k+P]==1&&in[j][k]==0)){
                       flag=false;
                       break;
                    }
                if(flag) G[2*i][2*j-1]=min(in[i][0],in[j][0]);
            }
        }
        memcpy(backup,G,sizeof(G));
        printf("%d ",Edmonds_Karp());
        int tol=0;
        for(int i=1;i<=N;i++){
            for(int j=1;j<=N;j++){
                if(G[2*i][2*j-1]<backup[2*i][2*j-1]){
                    Line[tol][0]=i;
                    Line[tol][1]=j;
                    Line[tol][2]=backup[2*i][2*j-1]-G[2*i][2*j-1];
                    tol++;
                }
            }
        }
        printf("%d\n",tol);
        for(int i=0;i<tol;i++)
            printf("%d %d %d\n",Line[i][0],Line[i][1],Line[i][2]);
    }

    return 0;
}