POJ 1325 Machine Schedule（二分匹配 最小點覆蓋）

標籤：poj   二分匹配   

題目連結：http://poj.org/problem?id=1325

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 5 100 1 11 1 22 1 33 1 44 2 15 2 26 2 37 2 48 3 39 4 30

Sample Output

3

Source

Beijing 2002

題目大意;有兩台機器A和B以及N個需要啟動並執行任務。每台機器有M種不同的模式，而每個任務都恰好在一台機器上運行。如果它在機器A上運行，則機器A需要設定為模式xi,如果它在機器B上運行，則機器A需要設定為模式yi。每台機器上的任務可以按照任意順序執行，但是每台機器每轉換一次模式需要重啟一次。請合理為每個任務安排一台機器併合理安排順序，使得機器重啟次數盡量少。二分圖的最小頂點覆蓋數=最大匹配數

代碼如下：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;const int MAXN = 117;int LN, RN;int g[MAXN][MAXN], linker[MAXN];bool used[MAXN];int dfs(int L){    for(int R = 0; R < RN; R++)    {        if(g[L][R] && !used[R])        {            used[R] = true;            if(linker[R] == -1 || dfs(linker[R]))            {                linker[R] = L;                return 1;            }        }    }    return 0;}int hungary(){    int res = 0;    memset(linker,-1,sizeof(linker));    for(int L = 0; L < LN; L++)    {        memset(used,0,sizeof(used));        if(dfs(L))            res++;    }    return res;}int main(){    int i, L, R;    int res;    int k;    while(~scanf("%d",&LN) && LN)    {        scanf("%d%d",&RN,&k);        memset(g,0,sizeof(g));        for(int j = 0; j < k; j++)        {            scanf("%d%d%d",&i,&L,&R);            if(L > 0 && R > 0)                g[L][R] = 1;        }        res = hungary();        printf("%d\n",res);    }    return 0;}

POJ 1325 Machine Schedule（二分匹配 最小點覆蓋）