Server Load balancer 10817-Headmaster's Headache (status compression DP)

Server Load balancer 10817-Headmaster's Headache (status compression DP)

There are S (S <= 8) courses, each course requires at least two teachers to teach, it takes a certain amount of money to hire each teacher, each teacher can teach one or more courses. Existing M (M <= 20) in-service teachers, N (N <= 100) candidates, for in-service teachers, each must be hired, what is the minimum cost if at least two instructors are required for each course.

D [I] [j] indicates the minimum cost when the status of the former I teacher is j. j indicates the number of S-bits in triplicate, if the value is 0, the course u is taught by no teacher. If the value is 1, the course u is taught by one teacher. If the value is 2, the course u is taught by two teachers. The final answer is d [N] [3 ^ S-1].

When calculating d [I] [j], consider whether to hire the I-th teacher to complete the status transfer. If hired, try to make full use of the Course taught by this teacher, for each course taught by this teacher, if the current j course corresponds to a three-digit number greater than 0, then minus 1, otherwise it will not change.

State transition equation: d [I] [j] = max {d [I-1] [j], d [I-1] [v] + cost [I]} (v is the State after subtracting the course that can be subtracted)

#include
 
  #include
  
   #include
   
    char a[300];int b[105][10];int c[10];int d[105][7300];int co[105];int basethre[7100][10];int basetco[10];int main(void){int i,j,u,m,n,p,q,lo,sum,sump,minp,OK;q=1;for(i=1;i<=8;i++){basetco[i]=q;q=q*3;}for(i=0;i
    
     ='0')&&(a[j]<='9')){sump=0;while((a[j]>='0')&&(a[j]<='9')){sump=sump*10+a[j]-'0';j++;}sum=sum+sump;break;}}for(u=j;u
     
      ='0')&&(a[u]<='9')&&(c[a[u]-'0']<2)){c[a[u]-'0']++;}}}for(i=1;i<=m;i++){gets(a);lo=strlen(a);for(j=0;j
      
       ='0')&&(a[j]<='9')){sump=0;while((a[j]>='0')&&(a[j]<='9')){sump=sump*10+a[j]-'0';j++;}co[i]=sump;break;}}for(u=j;u
       
        ='0')&&(a[u]<='9')){b[i][a[u]-'0']=1;}}}d[0][0]=sum;for(j=1;j<=q;j++){OK=1;for(u=1;u<=p;u++){if(basethre[j][u]>c[u]){OK=0;break;}}if(OK==1){d[0][j]=sum;}else{d[0][j]=1000000000;}}for(i=1;i<=m;i++){d[i][0]=sum;for(j=1;j<=q;j++){minp=d[i-1][j];sump=0;for(u=1;u<=p;u++){if((basethre[j][u]>0)&&(b[i][u]==1)){sump=sump+basetco[u];}}if(minp>d[i-1][j-sump]+co[i]){minp=d[i-1][j-sump]+co[i];}d[i][j]=minp;}}printf("%d\n",d[m][q]);for(i=0;i<=10;i++){c[i]=0;}memset(b,0,sizeof(b));scanf("%d%d%d",&p,&n,&m);getchar();}return 0;}