Relocation POJ - 2923 (01 shaped pressure backpack + dp)

Relocation

Links:  POJ - 2923 

Title effect: to give you two cars, each car has a rated load c1, c2, give you n items (n <= 10) of the weight Wi, each time with two round-trip transportation vehicles, the minimum number of requirements.

Idea: to see a small number n, first think of violence, can enumerate each case of loading, to determine whether the establishment of the situation prior to loading the state of compression, 0 for not installed, 1 to wear, to state. violence then readily occur to be enumerated with a backpack 01 determines whether the program is feasible, is determined as follows: add the total weight of the selected item, then c1 maximum weight capacity of the vehicle 01 can hold the backpack request, if the car c2 weight> -c1 total weight of the car can hold true representative of the maximum weight is returned possible. Next, the state of the article is dp, the state 0 to the state (1 << n) -1, each transfer means enumerate viable car program, the state transition equation is as follows:

dp[i|sta[j]]=min(dp[i|sta[j]],dp[i]+1);

sta [j] is the program loading enumeration, i | sta [j] is the assembled state after loading.

Code:

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <algorithm>
 4 #include<iostream>
 5 #include<cmath>
 6 #define INF 0x3f3f3f3f
 7 #define MAXN 100000
 8 using namespace std;
 9 typedef long long ll;
10 int f[105],dp[1<<12],w[MAXN],sta[MAXN],c1,c2,n;
11 bool check(int st)
12 {
13     memset(f,0,sizeof(f));
14     int sum=0;
15     for(int i=0;i<n;i++)
16     {
17         if(st&(1<<i))
18         {
19             sum+=w[i];
20             if(sum>c1+c2)continue;
21             for(int j=c1;j>=w[i];j--)
22             f[j]=max(f[j],f[j-w[i]]+w[i]);
23         }
24     }
25     if(c2>=sum-f[c1])return true;
26     return false;
27 }
28 int main()
29 {
30     ios::sync_with_stdio(false);
31     int cas;
32     cin>>cas;
33     for(int ca=1; ca<=cas; ca++)
34     {
35         memset(dp,INF,sizeof(dp));
36         dp[0]=0;
37         cin>>n>>c1>>c2;
38         c1 > c2 ? c1 ^= (c2 ^= (c1 ^= c2)) : 0;
39         for(int i=0;i<n;i++)
40             cin>>w[i];
41         int cnt=0;
42         for(int i=0;i<(1<<n);i++)
43             if(check(i))sta[cnt++]=i;
44         for(int i=0;i<(1<<n);i++)
45         {
46             for(int j=0;j<cnt;j++)
47             dp[i|sta[j]]=min(dp[i|sta[j]],dp[i]+1);
48         }
49         cout<<"Scenario #"<<ca<<":"<<endl;
50         cout<<dp[(1<<n)-1]<<endl;
51         if(ca<cas)cout<<endl;
52     }
53     return 0;
54 }

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