个人心得：二分图啥的一点都不知道，上网借鉴了下，请参考http://blog.csdn.net/thundermrbird/article/details/52231639
加上自己的了解，二分图就是将图形拆分成俩半，俩边的点通过边界相连接。匹配就是不存在俩条边存在同一顶点，就是一个顶点最多连接一条边。
匈牙利算法就是用增广路进行更新，仔细一想确实如此，如果存在这样的途径那么必然可以用亦或就行维护更新，细节看参考。
不过碍于自己图论太low，还是无法运用的好，比如这题，就是最小覆盖点，关于最小覆盖点等于最大匹配数还没搞明白。
总的来说这就是给我提供一个二分最大匹配的模板吧，算法细节方面有时间还是要去学的，不能单纯为了做题而学习算法！
励志语：没什么能够打倒你，能打倒你的只有你的自卑和懦弱。前方大路终究变得宽阔，加油。
题目：
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.
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.
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
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 #include<vector> 6 #include<queue> 7 #include<string> 8 #include<algorithm> 9 using namespace std; 10 const int inf=0x3f3f3f3f; 11 const int maxa=105; 12 int g[maxa][maxa]; 13 int book[maxa]; 14 int link[maxa]; 15 int n,m,k; 16 int dfs(int u){ 17 int v; 18 for(v=0;v<m;v++) 19 if(g[u][v]&&!book[v]) 20 { 21 book[v]=1; 22 if(link[v]==-1||dfs(link[v])) 23 { 24 link[v]=u; 25 return 1; 26 } 27 } 28 return 0; 29 } 30 int hungary() 31 { 32 int res=0; 33 int u; 34 memset(link,-1,sizeof(link)); 35 for(int i=0;i<n;i++) 36 { 37 memset(book,0,sizeof(book)); 38 if(dfs(i)) res++; 39 } 40 return res; 41 } 42 int main(){ 43 while(cin>>n){ 44 if(n==0) break; 45 memset(g,0,sizeof(g)); 46 cin>>m>>k; 47 int x,y,z; 48 while(k--){ 49 cin>>x>>y>>z; 50 if(y>0&&z>0) g[y][z]=1; 51 } 52 cout<<hungary()<<endl; 53 54 } 55 56 return 0; 57 }
