Machine Schedule
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5316 Accepted Submission(s): 2631
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.
The input will be terminated by a line containing a single zero.
转自：http://blog.csdn.net/vsooda/article/details/7447550
题目大意：
有两台机器A和B以及N个需要运行的任务。每台机器有M种不同的模式，而每个任务都恰好在一台机器上运行。
如果它在机器A上运行，则机器A需要设置为模式xi,如果它在机器B上运行，则机器A需要设置为模式yi。
每台机器上的任务可以按照任意顺序执行，但是每台机器每转换一次模式需要重启一次。请合理为每个任务安排一台机器并合理安排顺序，使得机器重启次数尽量少。
在二分图中求最少的点，让每条边都至少和其中的一个点关联，这就是：“二分图的最小顶点覆盖”求最少的点，让每条边都至少和其中的一个点关联，这就是
二分图的最小顶点覆盖数=最大匹配数
本题就是求最小顶点覆盖数的。
我就是没弄懂题意的..
//0MS 308K 989B G++
#include<iostream>
#include<vector>
#define N 105
using namespace std;
vector<int>V[N];
int vis[N];
int match[N];
int n,m;
int dfs(int x)
{
for(int i=;i<V[x].size();i++){
int v=V[x][i];
if(!vis[v]){
vis[v]=;
if(match[v]==- || dfs(match[v])){
match[v]=x;
return ;
}
}
}
return ;
}
int hungary()
{
memset(match,-,sizeof(match));
int ret=;
for(int i=;i<n;i++){
memset(vis,,sizeof(vis));
ret+=dfs(i);
}
return ret;
}
int main(void)
{
int a,b,c,k;
while(scanf("%d",&n)!=EOF && n)
{
scanf("%d%d",&m,&k);
int tn=n>m?n:m;
for(int i=;i<=n;i++)
V[i].clear();
for(int i=;i<k;i++){
scanf("%d%d%d",&c,&a,&b);
if(a && b)
V[a].push_back(b);
}
printf("%d\n",hungary());
}
return ;
}