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解题报告
http://blog.csdn.net/juncoder/article/details/38147135
题目传送门(POJ)
题目传送门(HDU)
题意:
A机器有n个模式,B机器有m个模式,每个作业可以在任何机器的特定模式下工作,转换模式需要耗时,求最小耗时
思路:
把AB两机器的模式当成二分图顶点,模式之间的连线就是某个作业可以在该两个模式下工作,就转换成求最小点覆盖,用最少的点覆盖最多的边。
最小点覆盖=最大匹配
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
int mmap[220][220],n,m,k,vis[220],pre[220];
int dfs(int x)
{int i;for(i=n;i<n+m;i++){if(!vis[i]&&mmap[x][i]){vis[i]=1;if(pre[i]==-1||dfs(pre[i])){pre[i]=x;return 1;}}}return 0;
}
int main()
{int i,j,a,x,y;while(cin>>n>>m>>k){if(!n)break;memset(pre,-1,sizeof(pre));memset(mmap,0,sizeof(mmap));for(i=0;i<k;i++){cin>>a>>x>>y;if(x*y!=0)mmap[x][y+n]=1;}int ans=0;for(i=0;i<n;i++){memset(vis,0,sizeof(vis));ans+=dfs(i);}cout<<ans<<endl;}
}
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 11631 | Accepted: 4952 |
Description
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 will be terminated by a line containing a single zero.
Output
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
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