poj 1325 Machine Schedule

Time Limit: 1000 MS Memory Limit: 10000 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

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  ///n，m，k两台机器人  机器人A可完成n项工作  机器人B可完成m项工作  有K项工作
0 1 1  i,x,y  工作i可由x,y完成   以下同理
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

第一道二分图最大匹配题  类模板~~

思路：构造二分图 A的n个节点 B的m个节点 分别看成图的两个集合的顶点   如果可以在A和B上完成 则A B之间连边  则构成了二分图
     本题求最小点覆盖集的问题  即最少的顶点覆盖全部的边（这样开关机器的次数就最少  因为为每一个点看成了机器）  转换成了求二分图的最大匹配问题  二分图的点覆盖数==匹配数

#include <iostream>

#include <string.h>

#include <stdio.h>

using namespace std;

#define maxn 105

int nx,ny,jobnum;  ///机器A  机器B  工作数

int g[maxn][maxn];  ///所构造的二分图

int ans;  ///最大匹配数

int sx[maxn],sy[maxn]; ///path函数中DFS中用来标明顶点访问状态的数组

int cx[maxn],cy[maxn];  ///匹配情况cx[i] x集合中顶点i匹配y集合顶点cx[i]~~

int path(int u)

{

    sx[u]=; ///标记访问过

    int v;  ///变量

    for(v=; v<=ny; v++)  ///一次访问y集合中的顶点

 ///以B集合为遍历条件  使其与集合A相对应
    {

        if((g[u][v]>)&&(!sy[v])) ///有连接&&v未访问过

        {

            sy[v]=; ///标记

            if(path(cy[v])||(!cy[v])) ///v已经匹配了 但可以在v找到可增广路

            {

                ///回退过程中修改可增广路的值  使匹配数加1

                cx[u]=v; ///v匹配给u,u匹配给iv

                cy[v]=u;

                return ;

            }

        }

    }

    return ;

}

int solve()

{

    ans=;

    int i;

    memset(cx,,sizeof(cx));  ///

    memset(cy,,sizeof(cy));

    for(i=; i<=nx; i++)

///以A集合为标准 进行遍历
      {

         if(!cx[i])

        {

            memset(sx,,sizeof(sx));

            memset(sy,,sizeof(sy));

            ans+=path(i); ///可增广路扩充的

        }

    }

 }

int main()

{

    int m,i,j;  ///i j 对应m工作

    while(scanf("%d",&nx)) ///A队nx人 B队ny人 jobnum份工作

    {

        if(nx==)

        break;

        scanf("%d%d",&ny,&jobnum);

        memset(g,,sizeof(g));  ///初始没有边相连

        for(int k=; k<jobnum; k++)

        {

            scanf("%d%d%d",&m,&i,&j);

            g[i][j]=;  ///构二分图 等于1表示i  j相连

        }

        solve();

        printf("%d\n",ans);

    }

    return ;

}

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