poj2923 Relocation（状态压缩+背包）

最新推荐文章于 2020-03-20 16:46:13 发布

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本博客讨论了如何在给定车辆载重限制的情况下，最小化将家具从旧居搬至新居所需的运输次数。通过状态转移算法，计算出最优的搬运方案。

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Relocation

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2100		Accepted: 871

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:

At their old place, they will put furniture on both cars.
Then, they will drive to their new place with the two cars and carry the furniture upstairs.
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 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 ≤ C_i ≤ 100) andn is the number of pieces of furniture (1 ≤ n ≤ 10). The following line will containn 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:”, 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

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/*
题目大意：有N个物品用两辆车运输，c1和c2问最少多少次能运走所有东西
难点：状态转化
Time：2024-8-11 9:54
*/
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int MAX=1<<15;
const int INF=1<<30;
int N,c1,c2;
int w[1000],dp[MAX],hash[MAX];//hash数组要开足够大，存很多分解后的数据
bool vis[MAX]; 
bool check(int x)
{
	int sum=0;
	memset(vis,0,sizeof(vis));
	 vis[0]=1;
	//把 0一定要置为 1，在下边的时候才能标记掉检索到的状态 
	for(int i=0;i<N;i++){
		if(x&(1<<i)){ //如果 其中的状态有小于 等于 x 的，则可以拉走 
			sum+=w[i]; 
			for(int j=c1-w[i];j>=0;j--){
				//w[i]能拉走，那么其中比 c1-w[i]小的 并且状态存在的也可以拉走 
				if(vis[j])
				vis[j+w[i]]=1;//所以标记掉 
			}
		} 
	}	
	for(int j=c1;j>=0;j--){
		//因为 c1比c2小，c1能拉走的，c2一定能拉走 所以 只需要检索小于c1的就行 
		if(vis[j]&&(sum-j<=c2))//vis[j]表示存在这个状态并且剩下的c2能拉走 
		return 1;	
	}
	return 0;
			
}
int main()
{
	int T;
	int d=0;
	scanf("%d",&T);
	while(T--){
		memset(w,0,sizeof(w));
		scanf("%d%d%d",&N,&c1,&c2);
		if(c1>c2)swap(c1,c2);
		
		for(int i=0;i<N;i++)
		scanf("%d",&w[i]);
		int maxn=(1<<N)-1;
		int num=0;
		
		for(int i=1;i<=maxn;i++)
			if(check(i))//检索该状态是否能两辆车同时运走 
			hash[num++]=i;
		 /* 
		for(int i=0;i<num;i++)
		printf("%d ",hash[i]);
		printf("\n");*/
		for(int i=1;i<=maxn;i++)
			dp[i]=INF;
			
			dp[0]=0;
		for(int i=0;i<num;i++){
			for(int j=maxn-hash[i];j>=0;j--){
				//从 除了这个重量的状态以外的开始找 用减法刚好能去掉该状态 
				if(!(j&hash[i]))//如果某两个状态有不同的部分，
								//就可以计算这两个总的状态的步数  
				dp[j|hash[i]]=min(dp[j|hash[i]],dp[j]+1);
			}
		}
		printf("Scenario #%d:\n%d\n\n",++d,dp[maxn]);
	}
return 0;
}