hdu1150—Machine Schedule（二分图最大匹配）

题目链接：传送门

Machine Schedule

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9187    Accepted Submission(s): 4609

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 Output

  

   3
  

 

解题思路：将机器A和机器B的模式看成顶点，如果某个任务可以在A的mode_i或B的mode_j完成，则从Ai到Bj连一条边。要想完成所有工作，则需要满足所有边的某个端点对应的工作模式，而要想启动机器次数最少，则是求图中的最小点覆盖数。

对于二分图：最小点覆盖数 = 最大匹配数

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <stack>
#include <queue>

using namespace std;

typedef long long ll;
const int N = 1000100;
const int M = 1000;
const int INF = 0x3fffffff;
const double eps = 1e-8;
const double PI = acos(-1.0);

struct Edge{
    int node;
    Edge*next;
}m_edge[N];
int girl[M];
Edge*head[M];
int Flag[M],Ecnt,cnt;

void init()
{
    Ecnt = cnt = 0;
    fill( girl , girl+M , 0 );
    fill( head , head+M , (Edge*)0 );
}

//b对g有好感
void mkEdge( int b , int g )
{
    m_edge[Ecnt].node = g;
    m_edge[Ecnt].next = head[b];
    head[b] = m_edge+Ecnt++;
}

bool find( int x )
{
    for( Edge*p = head[x] ; p ; p = p->next ){
        int s = p->node;   //有好感的女生
        if( !Flag[s] ){
            Flag[s] = true; //该女生在本轮匹配中被访问
            if( girl[s] == 0 || find(girl[s]) ){
                //女生没有对象或者另外一个男生能把这个妹纸让给x男
                girl[s] = x;
                return true;
            }
        }
    }
    return false;
}

//构建二分图
void Build( int k )
{
    int a,b,c;
    for( int i = 0 ; i < k ; ++i ){
        scanf("%d%d%d",&c,&a,&b);
        mkEdge(a,b);
    }
}

void solve( int n )
{
    for( int i =  1 ; i <= n ; ++i ){
        fill( Flag , Flag+M , 0 );
        if( find(i) ) ++cnt;
    }
}

int main()
{
    int n,m,k;
    while( ~scanf("%d",&n)&&n ){
        scanf("%d%d",&m,&k);
        init();
        Build(k);
        solve(n);
        printf("%d\n",cnt);
    }
    return 0;
}