Headmaster's Headache UVA - 10817

UVA-10817

ans[i][s1][s2]表示考虑前i个人时，有至少1人教的科目集合为s1，有至少2人教的科目集合为s2时的最少工资
集合用一个数字表示，转换成二进制后从后面开始数第i位的状态(1/0)表示第i个科目的状态(满足/不满足某条件)
st[i]表示第i个人能教的课程集合,cost[i]表示第i个人的工资
ans[i][s1'][s2']=min(ans[i-1][s1'][s2'],ans[i-1][s1][s2]+cost[i])
s1'=s1|st[i]
s2'=s2|(s1&st[i])
压缩掉一维：
显然，s1'>=s1,s2'>=s2
因此,ans[s1'][s2']=min(ans[s1'][s2'],ans[s1][s2]+cost[i]) s1和s2从大到小

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 using namespace std;
 char tt[];
 int s,m,n,ans1,sa,sb;
 int ans[<<][<<];
 int st[],cost[];
 int main()
 {
     int i,t,point,maxj,j,j1;
     scanf("%d%d%d",&s,&m,&n);
     while(s!=)
     {
         memset(ans,0x3f,sizeof(ans));
         memset(st,,sizeof(st));
         cin.getline(tt,);
         ans1=;
         sa=;sb=;
         for(i=;i<=m;i++)
         {
             //memset(tt,'\0',sizeof(tt));
             cin.getline(tt,);
             point=;
             sscanf(tt,"%d",&t);
             ans1+=t;
             while(tt[point]<=''&&tt[point]>='')    ++point;
             while(tt[point]==' ')    ++point;
             while(point<strlen(tt))
             {
                 sscanf(tt+point,"%d",&t);
                 t=<<(t-);
                 sb|=sa&t;
                 sa|=t;
                 while(tt[point]<=''&&tt[point]>='')    ++point;
                 while(tt[point]==' ')    ++point;
             }
         }
         ans[sa][sb]=ans1;
         for(i=;i<=n;i++)
         {
             //memset(tt,'\0',sizeof(tt));
             cin.getline(tt,);
             point=;
             sscanf(tt,"%d",&cost[i]);
             while(tt[point]<=''&&tt[point]>='')    ++point;
             while(tt[point]==' ')    ++point;
             while(point<strlen(tt))
             {
                 sscanf(tt+point,"%d",&t);
                 st[i]|=<<(t-);
                 while(tt[point]<=''&&tt[point]>='')    ++point;
                 while(tt[point]==' ')    ++point;
             }
         }
         maxj=(<<s)-;
         for(i=;i<=n;i++)
             for(j=maxj;j>=;j--)
                 for(j1=maxj;j1>=;j1--)
                 {
                     sa=j|st[i];
                     sb=j1|(j&st[i]);
                     ans[sa][sb]=min(ans[sa][sb],ans[j][j1]+cost[i]);
                 }
         printf("%d\n",ans[maxj][maxj]);
         scanf("%d%d%d",&s,&m,&n);
     }
     return ;
 }