TYVJ p2018"nescafé26" Kitten Mountain climbing Problem Solving report

P2018 "nescafé26" Kitten Climbing

Time: 1000ms/Space: 131072kib/java class name: main background

Freda and Rainbow raised n kittens, and on this day, the kittens were going to climb mountains. After a long journey, the kittens climbed to the top of the mountain, but they did not want to walk down the hill again (Woo Goo >_<).

Describe

Freda and Rainbow had to pay for them to take the ropeway down the mountain. The cable car on the ropeway has a maximum bearing weight of W, while the weight of n kittens is C1, C2 ... CN. Of course, the weight of the kitten on each cable car cannot exceed W. Freda and Rainbow pay $1 for every cable car they hire, so they want to know at least how many dollars they need to pay to get the N kittens down the hill?

Input format

The first line contains two integers separated by a space, N and W.
The next n lines are an integer per line, where the integer of line i+1 represents the weight ci of the I kitten.

Output format

Output an integer that requires at least the minimum number of dollars, which is the minimum number of cable cars required.

Test Sample 1 input

5 1996
1
2
1994
12

in

Output

2

Note

For 100% of data, 1<=n<=18,1<=ci<=w<=10^8.
Nescafé26

—————————————————————— I'm a split line —————————————————————————————

Id-dfs topic.

? Enumerate the answers from 1~n, and iterate to deepen the search.? Each time the DFS is enumerated, enumerate each kitten on which cable car is placed.　　This will be tle, so let's add a pruning:? * The I kitten can only be placed on the front I cable car (no meaning on the back of the car).   This at first glance does not understand, another look but found-this is not the drawer principle??? qaq~ This is AC, the core code is not long.

1 /*2 PROBLEM:TYVJ 20183 OJ:TYVJ4 USER:S.B.S.5 time:169 Ms6 memory:1316 KB7 length:n/a8 */9#include <iostream>Ten#include <cstdio> One#include <cstring> A#include <cmath> -#include <algorithm> -#include <queue> the#include <cstdlib> -#include <iomanip> -#include <cassert> -#include <climits> +#include <functional> -#include <bitset> +#include <vector> A#include <list> at#include <map> - #defineMAXN 100001 - #defineF (i,j,k) for (int i=j;i<=k;i++) - #defineM (A, B) memset (A,b,sizeof (a)) - #defineFF (i,j,k) for (int i=j;i>=k;i--) - #defineINF 0x3f3f3f3f in #defineMAXM 1001 - #defineMoD 998244353 to //#define LOCAL + using namespacestd; - intRead () { the     intx=0, f=1;CharCh=GetChar (); *      while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();} $      while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; ch=GetChar ();}Panax Notoginseng     returnx*F; - } the intn,m,w; + intdep,ans=-1; A intC[MAXN],RE[MAXN]; theInlinevoidDfsintu) + { -     intb;BOOLflag=false; $     if(ans!=-1)return; $     if(u>N) { -ans=DEP; -         return; the     } -F (I,1, Min (U,DEP)) {Wuyi         if(W-re[i]<c[u])Continue; there[i]+=c[u];flag=true; -DFS (u+1); Wu         if(ans!=-1)return; -re[i]-=C[u]; About     } $     if(flag==false)return; - } - intMain () - { AStd::ios::sync_with_stdio (false); + #ifdef LOCAL theFreopen ("data.in","R", stdin); -Freopen ("Data.out","W", stdout); $     #endif theCin>>n>>W; theF (I,1, N) cin>>C[i]; the      for(dep=1;; dep++){ theDfs1); -         if(ans!=-1){ incout<<ans<<Endl; the              Break; the         } About     } the     return 0; the}

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TYVJ p2018"nescafé26" Kitten Mountain climbing Problem Solving report