每个求职者的pi, 对于每个求职者，要么选，要么不选，就是01背包问题。

对于s1，s2，可以根据当前枚举到的求职者课程即可，可推出下一个状态：

nextS1 = p[i] | s1,

nextS2 = (p[i] & s1) | s2

f[nextS1][nextS2] = min(f[nextS1][nextS2], f[s1][s2] + p[i])

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/**========================================== *  
 This is a solution for ACM/ICPC problem * *
   @problem: UVA 10817 - Headmaster's Headache *  
 @type:  dp *   @author: shuangde * 
  @blog: blog.csdn.net/shuangde800 *
   @email: zengshuangde@gmail.com *===========================================*
/#include#include#include#include#include#include#includeusing
 namespace std;typedef long long int64;
const int INF = 0x3f3f3f3f;
const double PI  = acos(-1.0);
const int MAXN = 150;
int courseNum, m, n, sum;
 int maxState;int f[1<<8][1<<8];
int p[MAXN], c[MAXN], cnt[10];
int dp(int st1, int st2){    memset(f, 0x3f, sizeof(f));  
  f[st1][st2] = sum;  
  for(int i=m+1;
 i<=n+m; ++i){   
     for(int s1=maxState; s1>=0; --s1){   
         for(int s2=maxState;
 s2>=0; --s2){                if(f[s1][s2] >= INF) continue;    
            int st1 = (p[i]|s1);         
       int st2 = (p[i]&s1) | s2;  
              f[st1][st2] = min(f[st1][st2], f[s1][s2]+c[i]);      
      }         }    }    return f[maxState][maxState];}int main(){    char str[1000];  
  while(~scanf("%d%d%d",
 &courseNum, &m, &n) && courseNum){            maxState = (1< 1) st2 |= (1< 
 
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