poj 2923 Relocation

Relocation

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1479		Accepted: 597

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

题意：
    两个车有容量a与b，n个物品，每个物品有一个体积，问最小多少次能够运完他们。

感想：第一次做状态压缩 很多位置都被坑了

1.掌握位运算的运算法优先级别很重要 
2.掌握基本位运算  
  (1).判断是否为0  if((S&1<<i)==0) 
  (2).将第i位置1  S|1<<(i-1)
  (3).将第i位置0  S&~(1<<(i-1))
3.要有状态的思想 每次是对状态进行操作 

思路：(状态压缩DP+背包)
先枚举选择若干个时的状态，
总状态量为1<<n，判断这些状态集合里的那些物品能否一次就
运走，如果能运走，那就把这个状态看成一个物品。预处理完能
从枚举中找到tot个物品，再用这tol个物品中没有交集
（也就是两个状态不能同时含有一个物品）的物品进
行01背包，每个物品的体积是state[i]，价值是1，求
包含n个物品的最少价值也就是dp[(1<<n)-1]（dp[i]表示状态i需要运的最少次数)。

状态转移方程:dp[j|k] = min(dp[j|k],dp[k]+1) (k为state[i,1<=j<=(1<<n)-1])。
算法复杂度O((2^N)*N)

代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <algorithm>
#define maxn 1030
using namespace std;

const int INF=0x3f3f3f3f;
int n,c1,c2,cnt;
int wei;
int a[15],dp[maxn];
int vis[maxn],state[maxn];

bool isok(int x)                  // 判断状态是否成立(是否一次能运走)
{
    int i,j,temp,sum;
    memset(vis,0,sizeof(vis));
    vis[0]=1;
    sum=0;
    for(i=0;i<n;i++)
    {
        if(x&(1<<i))            
        {
            temp=a[i+1];
            sum+=temp;
            for(j=c1;j>=temp;j--)          // 感觉也是01背包的思想
            {
                if(vis[j-temp]) vis[j]=1;
            }
        }
    }
    for(i=0;i<=c1;i++)
    {
        if(vis[i])
        {
            if(sum-i<=c2) return true;
        }
    }
    return false;
}
int minn(int x,int y)
{
    return x<y?x:y;
}
void ZeroOnepack(int weight,int val)
{
    int i;
    for(i=wei;i>=0;i--)                   // 容量为 (1<<n)-1
    {
        if(dp[i]>=INF) continue;          // 状态保证不冲突
        if((i&weight)==0)             
        {
            dp[i|weight]=minn(dp[i|weight],dp[i]+val);   // 如果能一次搬走 就一次搬走 不行的话再加一次
        }
    }
}
int main()
{
    int i,j,t,k;
    scanf("%d",&t);
    for(k=1; k<=t; k++)
    {
        scanf("%d%d%d",&n,&c1,&c2);
        for(i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
        }
        cnt=0;
        wei=(1<<n)-1;
        for(i=1;i<=wei;i++)                  // 预处理
        {
            if(isok(i)) state[++cnt]=i;
        }
        memset(dp,0x3f,sizeof(dp));
        dp[0]=0;
        for(i=1;i<=cnt;i++)                  // 01背包  状态为物品
        {
            ZeroOnepack(state[i],1);
        }
        printf("Scenario #%d:\n%d\n\n",k,dp[wei]);
    }
    return 0;
}