ACM Computer Factory【网络流dinic+题意理解】

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,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 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
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

题目大意：
根据第一个样例来解释：每台电脑有3个零件，有4台安装电脑的机器，下面四行每行2*3+1个数，第一个数表示该台机器最多能生产多少台电脑，后面3个数字表示该台机器进入要求 （0表示不需要该零件，1表示需要该零件，2表示可有可无），最后三个数字是产出后电脑的规格（0表示没有该零件，1表示有该零件）；

解题思路：代码中写了注释，请直接代码，本人用的是dinic最大流算法：有疑问可以评论

#include<cstdio>
#include<cstring>
#include<string>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<map>
#include<vector>
#include<stack>
#define inf 0x3f3f3f3f
#include<queue>
using namespace std;
typedef long long ll;
const int N=55;

struct node//存网络流的边
{
    int u,v,ne,f;
}edge[N*N];
int head[N];
int t,s,dis[N],e;

struct mmp//每台机器的规格（就是机器的作用）
{
    int in[N],ou[N];//in[]数组事进入时的要求，ou[]是产出的要求
    int val;
}q[N];

void add(int a,int b,int c)//存边
{
    edge[e].u=a;//记得存一个起点，最后输出答案是用到
    edge[e].v=b;
    edge[e].ne=head[a];
    edge[e].f=c;
    head[a]=e++;
    edge[e].v=a;
    edge[e].u=b;
    edge[e].f=0;
    edge[e].ne=head[b];
    head[b]=e++;
}

int dinic_dfs(int now,int maxx)
{
    if(now==t) return maxx;
    int ret=0;
    for(int i=head[now];i!=-1;i=edge[i].ne)
    {
        int v=edge[i].v,f=edge[i].f;
        if(f&&dis[v]==dis[now]+1)
        {
            int mini=min(f,maxx-ret);
            f=dinic_dfs(v,mini);
            edge[i].f-=f;
            edge[i^1].f+=f;
            ret+=f;
            if(ret==maxx)return ret;
        }
    }
    return ret;
}

int dinic_bfs()
{
    queue<int>que;
    memset(dis,0,sizeof(dis));
    dis[s]=1;
    que.push(s);
    while(!que.empty())
    {
        int u=que.front();
        que.pop();
        if(u==t) return 1;
        for(int i=head[u];i!=-1;i=edge[i].ne)
        {
            int v=edge[i].v,f=edge[i].f;
            if(f&&(!dis[v]))
            {
                dis[v]=dis[u]+1;
                que.push(v);
            }
        }
    }
    return 0;
}

void dinic()//dinic模板
{
    int ans=0;
    while(dinic_bfs())
    {
        ans+=dinic_dfs(0,inf);
    }
    printf("%d ",ans);
}

int main()
{
    int p,n;
    while(~scanf("%d %d",&p,&n))
    {
        int vis1,vis2,maxx=0;
        e=0;s=0;t=n+1;
        memset(head,-1,sizeof(head));
        for(int i=1;i<=n;i++)
        {
            vis1=0;vis2=0;
            scanf("%d",&q[i].val);
            for(int j=1;j<=p;j++)//把进入要求全为0的直接建立边
            {
                scanf("%d",&q[i].in[j]);
                if(q[i].in[j]==1)
                    vis1=1;
            }
            for(int j=1;j<=p;j++)
            {
                scanf("%d",&q[i].ou[j]);//把产出要求全为1的直接建边
                if(q[i].ou[j]==0)
                    vis2=1;
            }
            if(vis1==0)
                add(0,i,q[i].val);
            if(vis2==0)
                add(i,n+1,q[i].val);
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(i==j) continue;
                vis1=0;
                for(int k=1;k<=p;k++)//然后把产出规格和进入规格相同的建边
                {
                    if(q[i].ou[k]!=q[j].in[k]&&q[j].in[k]!=2)
                       vis1=1;
                }
                if(vis1==0)
                    add(i,j,q[i].val);
            }
        }
        dinic();
        for(int i=0;i<e;i+=2)
        {
            if(edge[i].u==0||edge[i].v==n+1)
                continue;
            if(edge[i].f!=q[edge[i].u].val)
                maxx++;
        }
        printf("%d\n",maxx);
        for(int i=0;i<e;i+=2)
        {
            if(edge[i].u==0||edge[i].v==n+1)
                continue;
            if(edge[i].f!=q[edge[i].u].val)
                printf("%d %d %d\n",edge[i].u,edge[i].v,edge[i^1].f);
        }
    }
    return 0;
}