[2016-03-28] [HDU] [1074] [Doing Homework]

• Time: 2016-03-28-18:46:36
• Title Number: [2016-03-28][hdu][1074][doing homework]
• The main topic: given n the time period of the operation of the Department and the completion of times, each subject each more than one day to deduct a copy, the minimum deduction points
• Analysis: N only 15, binary enumeration, state compression, enumerate the status of each account completion, update the next state, find the minimum value

  
 

   
  

  
#include <cstring>
   
  

  
#include <cstdio>
   
  

  
using namespace std;
   
  

  
const int maxstu = 1 << 16;
   
  

  
struct Cource{
   
  

  
 char name[20];
   
  

  
 int d,c;
   
  

  
}a[16];
   
  

  
struct mstatus{
   
  

  
 int pre,cost,reduced,id;
   
  
Mstatus(inta= 0,intb= 0,intC= 0,intD= 0): Cost(a),reduced(b),Pre(C),ID(D){};
   
  

  
}dp[maxstu];
   
  

  
int vis[maxstu];
   
  

  
void print(int cur){
   
  

  
 if(dp[cur].pre == 0){
   
  

  
 printf("%s\n",a[ dp[cur].id ].name);
   
  

  
 return ;
   
  

  
 }
   
  

  
 print(dp[cur].pre);
   
  

  
 printf("%s\n",a[ dp[cur].id ].name);
   
  

  
}
   
  

  
int main(){
   
  

  
 int T;
   
  

  
 scanf("%d",&T);
   
  

  
 while(T--){
   
  

  
 int n;
   
  

  
 scanf("%d",&n);
   
  

  
 for(int i = 0;i < n;++i){
   
  

  
 scanf("%s%d%d",a[i].name,&a[i].d,&a[i].c);
   
  

  
 }
   
  

  
 memset(vis,0,sizeof(vis));
   
  

  
 dp[0] = mstatus(0,0,0,0);
   
  

  
  for   ( int   i  =    0  ;  i  <     ( 1    <<   n  )    -   1  ;++  i    
   
  

  
 for(int j = 0;j < n;++j){
   
  

  
 int cur = 1 << j;
   
  

  
 if((cur & i) == 0){
   
  

  
 int curcost = dp[i].cost + a[j].c;
   
  

  
  int   curreduced  =   DP  [ i  .  reduced  +     ( curcost  <=   a  [ j  .  d   0  :  curcost  -  a  [ j  .  d    
   
  

  
 int nxtstatus = cur | i;
   
  

  
 if(vis[nxtstatus]){
   
  

  
 if(curreduced < dp[nxtstatus].reduced){
   
  

  
 dp[nxtstatus] = mstatus(curcost,curreduced,i,j);
   
  

  
 }
   
  

  
 }else {
   
  

  
 vis[nxtstatus] = 1;
   
  

  
 dp[nxtstatus] = mstatus(curcost,curreduced,i,j);
   
  

  

   
  

  
 }
   
  

  
 }
   
  

  
 }
   
  

  
 }
   
  

  
 int ans = dp[(1 << n) - 1].reduced;
   
  

  
 printf("%d\n",ans);
   
  

  
 print((1 << n) - 1);
   
  

  
 }
   
  

  
 return 0;
   
  

  
}
  
 

From for notes (Wiz)

[2016-03-28] [HDU] [1074] [Doing Homework]