HDU 1059——Dividing【多重揹包 & 二進制優化】

題目傳送門

簡單多重揹包，體積爲硬幣數，價值爲幣值，可用二進制處理成01揹包求解，可用30對num進行優化。

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Problem Description

Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the same value, because then they could just split the collection in half. But unfortunately, some of the marbles are larger, or more beautiful than others. So, Marsha and Bill start by assigning a value, a natural number between one and six, to each marble. Now they want to divide the marbles so that each of them gets the same total value.
Unfortunately, they realize that it might be impossible to divide the marbles in this way (even if the total value of all marbles is even). For example, if there are one marble of value 1, one of value 3 and two of value 4, then they cannot be split into sets of equal value. So, they ask you to write a program that checks whether there is a fair partition of the marbles.

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Input

Each line in the input describes one collection of marbles to be divided. The lines consist of six non-negative integers n1, n2, …, n6, where ni is the number of marbles of value i. So, the example from above would be described by the input-line ``1 0 1 2 0 0’’. The maximum total number of marbles will be 20000.

The last line of the input file will be ``0 0 0 0 0 0’’; do not process this line.

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Output

For each colletcion, output Collection #k:'', where k is the number of the test case, and then eitherCan be divided.’’ or ``Can’t be divided.’’.

Output a blank line after each test case.

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Sample Input

1 0 1 2 0 0
1 0 0 0 1 1
0 0 0 0 0 0

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Sample Output

Collection #1:
Can’t be divided.

Collection #2:
Can be divided.

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題意：

有價值1、2、3、4、5、6的硬幣各多少個，判斷能否把他們分成價值相等的兩部分。

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分析：

• 簡單的多重揹包
• 二進制優化

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AC代碼：

#include <iostream>
#include <vector>
#include <utility>
#include <cstring>
#include <algorithm>
#include <map>
#include <queue>
#include <stack>
#include <cstdio>
#include <fstream>
#include <set>
using namespace std;
typedef long long ll;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#define INF 0x3f3f3f

const int MAX_N = 50000 + 7;
const int MAX_W = 120005 + 7;

int dp[MAX_W];
int w[MAX_N];
int main() {
	int s[7];
	int count = 0;
	while (cin >> s[1] >> s[2] >> s[3] >> s[4] >> s[5] >> s[6], s[1] || s[2] || s[3] || s[4] || s[5] || s[6]) {
		memset(dp, 0, sizeof dp);
		int k = 0;
		int sum = 0;
		for (int i = 1; i <= 6; i++) {
			sum += s[i] * i;
			for (int a = 1; a <= s[i]; a *= 2) {
				w[k++] = i * a;
				s[i] -= a;
			}
			if (s[i] > 0)
				w[k++] = i * s[i];
		}

		for (int i = 1; i < k; i++)
			for (int j = sum / 2; j >= w[i]; j--) {
				dp[j] = max(dp[j - w[i]] + w[i], dp[j]);
			}
	
		cout << "Collection #" << ++count << ":" << endl;
		if (2 * dp[sum / 2] == sum)
			cout << "Can be divided." << endl;
		else
			cout << "Can't be divided." << endl;
		cout << endl;
	}
	return 0;
}