poj-1325-Machine Schedule

poj-1325-Machine Schedule

Machine Schedule

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 15834		Accepted: 6776

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

Source

Beijing 2002

17957182

1325

Accepted

4392K

79MS

G++

882B

2017-12-16 21:03:57

两台机器，分别有N种和M种状态，对于每一项任务来说都可以由两台机器来完成。所以，现在只需要找出最小的状态数量就行了。

只需要找Machine 1 的最少转态数量就行了。因为所有的任务都可以由 Machine 1 来完成。（使用贪心，尽可能所有任务由machine1完成，完成不了的交给 machine 2） 

可以看出是： 二分图匹配。使用匈牙利算法可以解决。

（开始提交出现 wrong answer， 因为问的是需要改变的最小状态次数，最初状态是 mode_0 状态，所以对于 任务可以在 mode_0 的话，可以跳过不考虑。）

// poj-1325 
#include <cstdio> 
#include <cstring> 
const int MAXN = 1000 + 10; 

int n, m, k, mp[MAXN][MAXN], vis[MAXN], match[MAXN], len[MAXN]; 

bool dfs(int x){
	for(int i=0; i<len[x]; ++i){
		int y = mp[x][i]; 
		if(vis[y] == 0){
			vis[y] = 1; 
			if(match[y] < 0 || dfs(match[y])){
				match[y] = x; 
				return true; 
			} 
		} 
	}
	return false; 
}

int main(){
	freopen("in.txt", "r", stdin); 

	int id, x, y, ans; 
	while(scanf("%d", &n) != EOF){
		if(n == 0){ break; } 
		scanf("%d %d", &m, &k); 
		memset(len, 0, sizeof(len)); 
		memset(mp, 0, sizeof(mp)); 

		for(int i=0; i<k; ++i){
			scanf("%d %d %d", &id, &x, &y); 
			if( x == 0 || y == 0){
				continue; 
			}
			mp[ x ][ len[x] ] = y; 
			++len[x]; 
		}

		memset(match, -1, sizeof(match)); 
		ans = 0; 
		for(int i=1; i<=n; ++i){
			memset(vis, 0, sizeof(vis)); 
			if(dfs(i)){
				++ans; 
			}
		} 
		printf("%d
", ans ); 
	}
	return 0; 
}