The number of a pit to put more carrots, seeking a minimum pit

Topic links: https://ac.nowcoder.com/acm/contest/1014/A

Title Description

Freda and rainbow reared N kittens, this day, the kittens go hiking. Has passed away, the cat finally climbed the hill, but they do not want to weary and walk down the mountain (woo goo> _ <).

Enter a description:

Freda and rainbow had to spend money to make them sit down cable. The maximum loading on the cable car cableway weight is W, the weight of the N kittens are  $CNC_1, C_2 \ DOTS C_N C . 1 , C 2 ... C N . Of course, the weight of the cat on the cable car and not exceed per W. Each of renting a cable car, Freda and rainbow have to pay $ 100, so they want to know, how many dollars to pay a minimum of N these kittens are transported down the mountain?$

Output Description:

The first line contains two space-separated integers, N and W. 
Next, an integer N lines, where the i + 1 row represents an integer of i-th kittens wt  $CiC_i C i .$

Example 1

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2 

　　prime ring with a considerable carrot a pit, the pit is a kitten climbing plurality radish, determined by the minimum number of pits. And prime ring is searching for each pit, the first step to see which several pit put; 
kitten climbing is traversed to find pit for him for the first step by the number, there is no need for the re-establishment of a pit. 
    It is worth mentioning that the written if (cnt> res) in the minus branch by submitting a written test is much slower than if (cnt> = res), former time in about 300ms, which is 4ms; Thus, pruning it important, 
proper pruning is more important. 

AC code is as follows:

. 1 #include <cstdio>
 2 #include <algorithm>
 . 3  #define Max 0x3f3f3f3f
 . 4  the using  namespace STD;
 . 5  
. 6  int n-, W, RES, ANS [ 20 is ], PIT [ 25 ];    
 . 7  
. 8  BOOL CMP ( int A, int B) {
 . 9      return a> B;
 10  }
 . 11  // STEP current process is the first of several cat, cnt processing the plurality of step-1 cat car 
12 is  void DFS ( int STEP, int CNT) {
 13 is      IF (CNT > =RES)
 14          return ;
 15      IF (step n-== + . 1 ) {
 16          RES = min (RES, CNT);
 . 17          return ;
 18 is      }
 . 19      
20 is      // car in the car looking for the conventional step of cats 
21      for ( int I = . 1 ; I <= CNT; I ++ ) {
 22 is          IF (PIT [I] + ANS [STEP] <= W) {
 23 is              PIT [I] + = ANS [STEP];
 24              DFS (STEP + . 1 , CNT);
 25              PIT [I] - = ANS [STEP];
 26 is         }
 27      }
 28      // find a suitable vehicle in a conventional car new one, and into which the cat 
29      PIT [CNT + . 1 ] = ANS [STEP];
 30      DFS (+ STEP . 1 , CNT + . 1 );
 31 is  }
 32  
33 is  int main ( void )
 34 is  {
 35      Scanf ( " % D% D " , & n-, & W);
 36      for ( int I = . 1 ; I <= n-; I ++ )
 37 [          Scanf ( " % D " , &ans[i]);
38     sort(ans+1,ans+1+n,cmp);
39     res = Max;
40     dfs(1,1);
41     printf("%d\n",res);
42 
43     return 0;
44 }