poj 2923

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本想做点简单点的DP练练手。。然后一不小心选到了状压DP。。。然后发现自己的DP真是弱的一比。。。

把几份物体被2车载一次的所有方案暴力出来,记录体积,然后就像01背包一样转移即可。。

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<queue>#define inc(i,l,r) for(int i=l;i<=r;i++)#define dec(i,l,r) for(int i=l;i>=r;i--)#define link(x) for(edge *j=h[x];j;j=j->next)#define inf 1e9#define mem(a) memset(a,0,sizeof(a))#define ll long long#define succ(x) (1<<x)#define lowbit(x) (x&&(-x))#define NM 2005using namespace std;ll read(){ll x=0,f=1;char ch=getchar();while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}while(isdigit(ch))x=x*10+ch-'0',ch=getchar();return f*x;}bool b[NM],c[NM];int a[20],t,s,n,T,m,tot,t1,t2,f[NM],d[NM];void dfs(int x){if(x==n+1){if(tot<=t1)b[s]=true;if(tot<=t2)c[s]=true;return;}t*=2;dfs(x+1);t/=2;s+=t;t*=2;tot+=a[x];dfs(x+1);t/=2;s-=t;tot-=a[x];}int main(){//freopen("data.in","r",stdin);T=read();inc(p,1,T){mem(a);mem(b);mem(c);mem(d);mem(f);tot=0;s=0;printf("Scenario #%d:\n",p);n=read();t1=read();t2=read();inc(i,1,n)a[i]=read();dfs(t=1);m=succ(n)-1;tot=0;//inc(i,1,m)if(b[i])printf("%d ",i);printf("\n");//inc(i,1,m)if(c[i])printf("%d ",i);printf("\n");dec(i,m,1)inc(j,0,i)if((i|j)==i&&b[j]&&c[i-j])f[++tot]=i,j=i;//printf("%d\n",tot);//inc(i,1,tot)printf("%d ",f[i]);inc(i,1,m)d[i]=inf;inc(i,1,tot)dec(j,m,f[i])if(f[i]==j)d[j]=1;elseif((f[i]&j)==f[i]&&d[j^f[i]])d[j]=min(d[j],d[j^f[i]]+1);printf("%d\n\n",d[m]);}return 0;}

Relocation
Time Limit: 1000MS
Memory Limit: 65536KTotal Submissions: 3544
Accepted: 1458

Description

Emma and Eric are moving to their new house they bought after returning from their honeymoon. Fortunately, they have a few friends helping them relocate. To move the furniture, they only have two compact cars, which complicates everything a bit. Since the furniture does not fit into the cars, Eric wants to put them on top of the cars. However, both cars only support a certain weight on their roof, so they will have to do several trips to transport everything. The schedule for the move is planed like this:

  1. At their old place, they will put furniture on both cars.
  2. Then, they will drive to their new place with the two cars and carry the furniture upstairs.
  3. Finally, everybody will return to their old place and the process continues until everything is moved to the new place.

Note, that the group is always staying together so that they can have more fun and nobody feels lonely. Since the distance between the houses is quite large, Eric wants to make as few trips as possible.

Given the weights wi of each individual piece of furniture and the capacitiesC1 and C2 of the two cars, how many trips to the new house does the party have to make to move all the furniture? If a car has capacityC, the sum of the weights of all the furniture it loads for one trip can be at mostC.

Input

The first line contains the number of scenarios. Each scenario consists of one line containing three numbersn, C1 and C2. C1 andC2 are the capacities of the cars (1 ≤ Ci ≤ 100) andn is the number of pieces of furniture (1 ≤ n ≤ 10). The following line will containn integers w1, …, wn, the weights of the furniture (1 ≤wi ≤ 100). It is guaranteed that each piece of furniture can be loaded by at least one of the two cars.

Output

The output for every scenario begins with a line containing “Scenario #i:”, wherei is the number of the scenario starting at 1. Then print a single line with the number of trips to the new house they have to make to move all the furniture. Terminate each scenario with a blank line.

Sample Input

26 12 133 9 13 3 10 117 1 1001 2 33 50 50 67 98

Sample Output

Scenario #1:2Scenario #2:3

Source

TUD Programming Contest 2006, Darmstadt, Germany

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