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
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 using namespace std; 5 char tt[500]; 6 int s,m,n,ans1,sa,sb; 7 int ans[1<<8][1<<8]; 8 int st[110],cost[110]; 9 int main() 10 { 11 int i,t,point,maxj,j,j1; 12 scanf("%d%d%d",&s,&m,&n); 13 while(s!=0) 14 { 15 memset(ans,0x3f,sizeof(ans)); 16 memset(st,0,sizeof(st)); 17 cin.getline(tt,500); 18 ans1=0; 19 sa=0;sb=0; 20 for(i=1;i<=m;i++) 21 { 22 //memset(tt,' ',sizeof(tt)); 23 cin.getline(tt,500); 24 point=0; 25 sscanf(tt,"%d",&t); 26 ans1+=t; 27 while(tt[point]<='9'&&tt[point]>='0') ++point; 28 while(tt[point]==' ') ++point; 29 while(point<strlen(tt)) 30 { 31 sscanf(tt+point,"%d",&t); 32 t=1<<(t-1); 33 sb|=sa&t; 34 sa|=t; 35 while(tt[point]<='9'&&tt[point]>='0') ++point; 36 while(tt[point]==' ') ++point; 37 } 38 } 39 ans[sa][sb]=ans1; 40 for(i=1;i<=n;i++) 41 { 42 //memset(tt,' ',sizeof(tt)); 43 cin.getline(tt,500); 44 point=0; 45 sscanf(tt,"%d",&cost[i]); 46 while(tt[point]<='9'&&tt[point]>='0') ++point; 47 while(tt[point]==' ') ++point; 48 while(point<strlen(tt)) 49 { 50 sscanf(tt+point,"%d",&t); 51 st[i]|=1<<(t-1); 52 while(tt[point]<='9'&&tt[point]>='0') ++point; 53 while(tt[point]==' ') ++point; 54 } 55 } 56 maxj=(1<<s)-1; 57 for(i=1;i<=n;i++) 58 for(j=maxj;j>=0;j--) 59 for(j1=maxj;j1>=0;j1--) 60 { 61 sa=j|st[i]; 62 sb=j1|(j&st[i]); 63 ans[sa][sb]=min(ans[sa][sb],ans[j][j1]+cost[i]); 64 } 65 printf("%d ",ans[maxj][maxj]); 66 scanf("%d%d%d",&s,&m,&n); 67 } 68 return 0; 69 }