poj 2923 Relocation(状态压缩+01背包)

Relocation

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 749		Accepted: 301

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 w_i of each individual piece of furniture and the capacities C₁ 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 capacity C, the sum of the weights of all the furniture it loads for one trip can be at most C.

Input

The first line contains the number of scenarios. Each scenario consists of one line containing three numbers n, C₁ and C₂. C₁ and C₂ are the capacities of the cars (1 ≤ C_i ≤ 100) and n is the number of pieces of furniture (1 ≤ n ≤ 10). The following line will contain n integers w₁, …, w_n, the weights of the furniture (1 ≤ w_i ≤ 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:”, where i 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

2
6 12 13
3 9 13 3 10 11
7 1 100
1 2 33 50 50 67 98

Sample Output

Scenario #1:
2

Scenario #2:
3

Source

TUD Programming Contest 2006, Darmstadt, Germany

题目： http://poj.org/problem?id=2923

分析：这题尽管数据看似很小，其实却非常麻烦，知道大概是DP，却无从下手，不过看到n<10 大概所有人都会想到状态压缩吧，没错，就是状态压缩，保存每一种状态是否满足能一次运走，写个判断函数，用01背包来判断这种状态是否可达，由于这些物品要装到两辆车上，也就是先01背包装第一辆车，剩下的只要判断第二辆车能否装下剩下的物品就行。。。接下来就是DP的核心，还是一个01背包，f[i]表示i状态最少用几次运，f[i]=min(f[j]+1) j|s[k]=i,s[k]为所有能一次运的状态。。。。

如果写一个判掉所有被覆盖的状态，这题应该能秒掉，不写的话要94ms

代码：

#include<cstdio>
#include<iostream>
using namespace std;
const int mm=1111;
int g[mm],f[mm],w[11];
bool v[111];
int i,j,n,c1,c2,t,s,cas=0;
bool ok(int x)
{
    int i,j,sum=0;
    for(i=0;i<=c1;++i)v[i]=0;
    v[0]=1;
    for(i=0;i<n;++i)
        if(x&(1<<i))
        {
            sum+=w[i];
            for(j=c1-w[i];j>=0;--j)
                if(v[j])v[j+w[i]]=1;
        }
    for(i=c1;i>=0;--i)
        if(v[i]&&sum-i<=c2)return 1;
    return 0;
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&c1,&c2);
        if(c1>c2)swap(c1,c2);
        for(i=0;i<n;++i)scanf("%d",&w[i]);
        for(s=0,i=1;i<1<<n;++i)
            if(ok(i))g[s++]=i;
        for(i=0;i<1<<n;++i)f[i]=1000000000;
        f[0]=0;
        for(i=0;i<s;++i)
            for(j=(1<<n)-1-g[i];j>=0;--j)
                if(!(j&g[i]))f[j|g[i]]=min(f[j|g[i]],f[j]+1);
        printf("Scenario #%d:\n%d\n\n",++cas,f[(1<<n)-1]);
    }
    return 0;
}