POJ 3436-ACM Computer Factory（最大流輸出路徑-Edmond-Karp演算法）

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 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,1D_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 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 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 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

Bold texts appearing in the sample sections are informative and do not form part of the actual data.

Northeastern Europe 2005, Far-Eastern Subregion

#include<iostream>
#include<cstdio>
#include<iomanip>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<map>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define INF 0x3f3f3f3f
#define MAXN 1010
struct Node
{
    int in[11];
    int out[11];
    int q;
} s[MAXN];
struct node
{
    int from,to,cap;
    node(int a,int b,int c):from(a),to(b),cap(c) {}
};
vector<node> res;//記錄網路流中機器之間的流量
int temp[MAXN][MAXN];
int capacity[MAXN][MAXN]; //記錄殘留網路的容量
int flow[MAXN];           //標記從源點到當前節點實際還剩多少流量可用
int pre[MAXN];            //標記在這條路徑上當前節點的前驅,同時標記該節點是否在佇列中
int n,p;                  //總結點數、組成部分數
queue<int> myqueue;
int BFS(int src,int des)
{
    int i;
    while(!myqueue.empty())       //佇列清空
        myqueue.pop();
    for(i=0; i<=n+1; ++i)
        pre[i]=-1;
    pre[src]=0;
    flow[src]= INF;
    myqueue.push(src);
    while(!myqueue.empty())
    {
        int index = myqueue.front();
        myqueue.pop();
        if(index == des)            //找到了增廣路徑
            break;
        for(i=0; i<=n+1; ++i)
        {
            if(i!=src && capacity[index][i]&& pre[i]==-1)
            {
                pre[i] = index; //記錄前驅
                flow[i] = min(capacity[index][i],flow[index]);   //關鍵：迭代的找到增量
                myqueue.push(i);
            }
        }
    }
    if(pre[des]==-1)      //殘留圖中不再存在增廣路徑
        return -1;
    else
        return flow[des];
}
int maxFlow(int src,int des)
{
    int increasement= 0;
    int sumflow = 0;
    while((increasement=BFS(src,des))!=-1)
    {
        int k = des;          //利用前驅尋找路徑
        while(k!=src)
        {
            int last = pre[k];
            capacity[last][k] -= increasement; //改變正向邊的容量
            capacity[k][last] += increasement; //改變反向邊的容量
            k = last;
        }
        sumflow += increasement;
    }
    return sumflow;
}

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("F:/cb/read.txt","r",stdin);
    //freopen("F:/cb/out.txt","w",stdout);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int i,j;
    while(cin>>p>>n)
    {
        int t=n;
        n*=2;
        memset(capacity,0,sizeof(capacity));
        memset(flow,0,sizeof(flow));
        for(i=1; i<=t; ++i)
        {
            cin>>s[i].q;
            capacity[i][i+t]=s[i].q;//
            bool b0=true,b1=true;
            for(j=0; j<p; ++j)
            {
                cin>>s[i].in[j];
                if(s[i].in[j]==1) b0=false;//存在1
            }
            for(j=0; j<p; ++j)
            {
                cin>>s[i].out[j];
                if(!s[i].out[j]) b1=false;//存在0
            }
            if(b0) capacity[0][i]=INF;//超級源點0
            if(b1) capacity[i+t][n+1]=INF;//超級匯點n+1
        }
        bool flag=true;
        for(i=1; i<=t; ++i)
            for(j=1; j<=t; ++j)
                if(i!=j)
                {
                    flag=true;
                    for(int k=0; k<p; ++k)
                        if(s[i].out[k]+s[j].in[k]==1)
                        {
                            flag=false;
                            break;
                        }
                    if(flag)
                        capacity[i+t][j]=INF;//機器之間
                }
        memcpy(temp,capacity,sizeof(temp));//複製capacity陣列到temp
        cout<<maxFlow(0,n+1)<<" ";
        for(int i=1; i<=t; i++)
        {
            for(int j=1; j<=t; j++)
            {
                if(i!=j)
                    if(capacity[t+i][j]<temp[t+i][j])//流量有減少說明需要使用該機器
                        res.push_back(node(i,j,temp[t+i][j]-capacity[t+i][j]));
            }
        }
        cout<<res.size()<<endl;
        for(int i=0; i<res.size(); i++)
            cout<<res[i].from<<" "<<res[i].to<<" "<<res[i].cap<<endl;
        res.clear();
    }
    return 0;
}
/*
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
*/