状态压缩DP——POJ 2923

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Exponentiation

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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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：

        有n件家具和两辆车（容量分别为c1, c2），某人要搬家，问至少需要几次才能把所有家具运完（两辆车必须同时行驶）。

思路：

        n很小，可以用二进制保存1 ~ (1<<n) - 1 种状态，每种状态表示几件家具，比如n = 7时。5这个状态，5用二进制表示为 0000101，可以用它表示为第5件家具和第7件家具被选上(或者从右边数起，表示第1和第3件家具)。之后可以枚举每种状态i，之后用两个循环就可完成。转移方程为：dp[i | j] = min{dp[i | j],  dp[i] + 1}; （j表示j状态能否一次被两辆车运走，其中i & j 要为空）。dp[x] 表示x状态至少需要几次才能拉走，所以dp[(1<<n) - 1] 就是答案。

        再来看怎样求一个状态x，使它能一次被两辆车运走 。先根据状态x的二进制中为1的位置进行对应家具求和sum，然后枚举x的子状态，同样据子状态的二进制中为1的位置进行对应家具求和sum1（表示此sum1要被两辆车中的某辆一次拉走），sum2 = sum - sum1。如果sum1和sum2都符合载重。则状态x能一次被两辆车运走。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 10
#define M (1<<N)
#define INF (1<<30)
#define MIN(x, y) ((x) < (y) ? (x) : (y))
#define SIZE ((1<<n)-1)
int a[N];
int dp[M];
int OK[M];
int c1, c2;

bool Judge(int x)
{
	int i, j, t = 0;
	int sum1 = 0, sum2 = 0, sum = 0;
	for(i=1; i<=x; i=(1<<t)){ /* 对x集合所包含的物品求和 */
		if(i & x) sum += a[t];
		t++;
	}
	if(sum > c1 + c2) return 0;
	for(i=1; i<=x; i++){ /* 枚举x的子集 */
		if((i & x) != i) continue; /* i不是x的子集 */
		sum1 = t = 0;
		for(j=1; j<=i; j=(1<<t)){
			if(j & i) sum1 += a[t];
			t++;
		}
		sum2 = sum - sum1;
		if(sum1 <= c1 && sum2 <= c2) return 1;
		if(sum2 <= c1 && sum1 <= c2) return 1;
	}
	return 0;
}

int main()
{
	//freopen("in.txt", "r", stdin);
	int T, n, w = 0;
	int i, j, ans;
	scanf("%d", &T);
	while(T--)
	{
		memset(OK, 0, sizeof(OK));
		scanf("%d%d%d", &n, &c1, &c2);
		for(i=0; i<n; i++) scanf("%d", a + i);
		for(i=1; i<M; i++) dp[i] = INF;

		for(i=1; i<=SIZE; i++)
			if(Judge(i)) OK[i] = dp[i] = 1;

		for(i=1; i<=SIZE; i++){
			for(j=1; j<=SIZE; j++){
				if(i & j) continue; /* 交集不为空 */
				if(dp[i] != INF && OK[j])
					dp[i|j] = MIN(dp[i|j], dp[i] + 1);
			}
		}

		printf("Scenario #%d:\n%d\n", ++w, dp[SIZE]);
		if(T) printf("\n");
	}
	return 0;
}