uva 10817 - Headmasters Headache ( 狀態壓縮dp)

每個求職者的pi, 對於每個求職者，要麼選，要麼不選，就是01背包問題。

對於s1，s2，可以根據當前枚舉到的求職者課程即可，可推出下一個狀態：

nextS1 = p[i] | s1,

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

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

/**========================================== *  
 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<iostream>#include<cstdio>#include<algorithm>#include<vector>#include<queue>#include<cmath>#include<cstring>using
 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<<courseNum) - 1;      

  sum = 0;   
     int st1=0, st2=0;    
    memset(cnt, 0, sizeof(cnt));     
           for(int i=1; i<=m+n; ++i){          
  scanf("%d", &c[i]);       
     gets(str);     
       p[i] = 0;        
    for(int j=0; j<strlen(str); ++j){                if(isdigit(str[j])){  
                  int num = str[j] - '0';     
               p[i] |= 1<<(num-1);      
              if(i <= m) ++cnt[num-1];          
      }             }            if(i <= m){                 sum += c[i];      
          st1 |= p[i];     
       }        }         for(int i=0; i<courseNum;
 ++i)            if(cnt[i] > 1) st2 |= (1<<i);    
    printf("%d\n", dp(st1, st2));  
  }    return 0;}