HDU 1150 Machine Schedule（最大流）

Problem Description

As we all know, machine scheduling is a very classical problem in computer science and has been studied for a very long history. Scheduling problems differ widely in the nature of the constraints that must be satisfied and the type of schedule desired. Here we consider a 2-machine scheduling problem.

There are two machines A and B. Machine A has n kinds of working modes, which is called mode_0, mode_1, …, mode_n-1, likewise machine B has m kinds of working modes, mode_0, mode_1, … , mode_m-1. At the beginning they are both work at mode_0.

For k jobs given, each of them can be processed in either one of the two machines in particular mode. For example, job 0 can either be processed in machine A at mode_3 or in machine B at mode_4, job 1 can either be processed in machine A at mode_2 or in machine B at mode_4, and so on. Thus, for job i, the constraint can be represent as a triple (i, x, y), which means it can be processed either in machine A at mode_x, or in machine B at mode_y.

Obviously, to accomplish all the jobs, we need to change the machine's working mode from time to time, but unfortunately, the machine's working mode can only be changed by restarting it manually. By changing the sequence of the jobs and assigning each job to a suitable machine, please write a program to minimize the times of restarting machines.

Input

The input file for this program consists of several configurations. The first line of one configuration contains three positive integers: n, m (n, m < 100) and k (k < 1000). The following k lines give the constrains of the k jobs, each line is a triple: i, x, y.

The input will be terminated by a line containing a single zero.

Output

The output should be one integer per line, which means the minimal times of restarting machine.

Sample Input

5 5 10

0 1 1 1 1 2 2 1 3 3 1 4 4 2 1 5 2 2 6 2 3 7 2 4 8 3 3 9 4 3 0

Sample Outpu

3

 题目大意：有两台机器，A 有 n 种工作状态，B 有 m 种工作状态，给出 k 个工作，每个工作可以在 A 的 x 状态下工作或者 B 的 y 状态下工作。每台机器切换工作状态需要重启，机器初始时都工作在 0 状态。求重排 k 个工作的执行顺序后，完成 k 个工作需要重启机器的最少次数。

可以将每个工作看作一条边，机器的状态看作结点，则给出的数据构成一个二分图，最少重启数可以看作是用最少的点使每条边至少与一个结点关联（即最小点覆盖），根据定理，二分图最小点覆盖该等于最大匹配，最大流即可解决。

需要注意的一点是机器的初始状态都为零，所以应当把包含状态 0 的工作放在最前面做，二分图建图时这些点不包含在里面。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
typedef struct node
{
    int from,to,cap,flow;
    node(int f = 0,int t = 0,int c = 0,int ff = 0):from(f),to(t),cap(c),flow(ff){}
}node;
const int INF = 1<<30;
const int maxn = 200 + 5;
int s,t,n,m;
vector<node> edge;
vector<int> G[maxn];
int cur[maxn],di[maxn];
bool vis[maxn];
void addedge(int from,int to,int cap,int flow)
{
    edge.push_back(node(from,to,cap,0));
    edge.push_back(node(to,from,0,0));
    int m = edge.size();
    G[from].push_back(m-2);
    G[to].push_back(m-1);
}
bool BFS()
{
    int x;
    memset(vis,false,sizeof(vis));
    queue<int> q;
    q.push(s);
    vis[s] = true;
    di[s] = 0;
    while(!q.empty())
    {
        x = q.front();
        q.pop();
        for(int i = 0;i < G[x].size(); ++i)
        {
            node &e = edge[G[x][i]];
            if(e.cap > e.flow && !vis[e.to])
            {
                di[e.to] = di[x] + 1;
                vis[e.to] = true;
                q.push(e.to);
            }
        }
    }
    return vis[t];
}
int DFS(int x,int a)
{
    int flow = 0,f;
    if(x == t || a == 0) return a;
    for(int &i = cur[x];i< G[x].size(); ++i)
    {
        node &e = edge[G[x][i]];
        if(di[e.to] == di[x] + 1 && (f = DFS(e.to,min(a,e.cap - e.flow))) > 0)
        {
            e.flow += f;
            edge[G[x][i]^1].flow -= f;
            flow += f;
            a -= f;
            if(a == 0) break;
        }
    }
    return flow;
}
int maxflow()
{
    int flow = 0;
    while(BFS())
    {
        memset(cur,0,sizeof(cur));
        flow += DFS(s,INF);
    }
    return flow;
}

bool vi[300];
int main()
{
    int x,y,k,tp;
    s = 0,t = 200;
    while(scanf("%d",&n) && n)
    {
        memset(vi,false,sizeof(vi));
        edge.clear();
        for(int i = 0;i < maxn; ++i) G[i].clear();
        scanf("%d %d",&m,&k);
        for(int i = 0;i < k; ++i)
        {
            scanf("%d %d %d",&tp,&x,&y);
            {
                if(!x || !y) continue;
                ++x,++y;
                if(!vi[x])
                {
                    vi[x] = true;
                    addedge(s,x,1,0);
                }
                if(!vi[y + 100])
                {
                    vi[y + 100] = true;
                    addedge(y + 100,t,1,0);
                }
                addedge(x,y + 100,1,0);
            }
        }
        cout << maxflow() << endl;
    }
    return 0;
}