POJ 1325 二分图最小点集覆盖

        一道貌似难题的水题，，不过我还是想了两个多小时。首先理解题意就理解了好久，，然后又想建图又想了好久，，最后发现建过图之后就是一道裸的二分图最小点集覆盖。。。。悲催。。。题目：

Machine Schedule

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 8033 Accepted: 3409

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

ac代码：

#include <iostream>#include <algorithm>#include <vector>#include <string.h>#include <cstdio>using namespace std;vector<int> ss[105];int flag[105],visted[105];bool dfs(int x){for(int i=0;i<ss[x].size();++i){if(!visted[ss[x][i]]){  visted[ss[x][i]]=1;  if(flag[ss[x][i]]==-1||dfs(flag[ss[x][i]])){    flag[ss[x][i]]=x;return true;  }}}return false;}int main(){  int n,m,k;  while(scanf("%d",&n)&&n){    scanf("%d%d",&m,&k);memset(flag,-1,sizeof(flag));memset(ss,0,sizeof(ss));int num,x,y;while(k--){  scanf("%d%d%d",&num,&x,&y);  if(!y)  continue;  ss[x].push_back(y);}int sum=0;for(int i=1;i<=n;++i){  memset(visted,0,sizeof(visted));  if(dfs(i))  sum++;}printf("%d\n",sum);  }  return 0;}