POJ1014-Dividing 多重背包+二进制划分

Dividing

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 74312		Accepted: 19441

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.

Input

Each line in the input file describes one collection of marbles to be divided. The lines contain six non-negative integers n1 , . . . , 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.

Output

For each collection, output "Collection #k:", where k is the number of the test case, and then either "Can be divided." or "Can't be divided.". 
Output a blank line after each test case.

链接

一、题意

        有价值为1~6的大理石若干，问这些大理石能否被划分为价值相等的两堆。

二、思路

        设总价值为sum，当sum为奇数时不能被划分，当sum为偶数时，该问题可以转换为是否能装满总额为sum/2的背包，即多重背包问题。因为每种价值的大理石可能很多，多以采用二进制划分减少复杂度。

三、代码

#include <iostream>
#include <sstream>
#include <iomanip>
#include <string>
#include <numeric>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <utility>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>

using namespace std;
typedef long long ll;
const int MAXN = 200100;
const int MOD7 = 1000000007;
const int MOD9 = 1000000009;
const int INF = 2000000000;//0x7fffffff
const double EPS = 1e-9;
const double PI = 3.14159265358979;
const int dir_4r[] = { -1, 1, 0, 0 };
const int dir_4c[] = { 0, 0, -1, 1 };
const int dir_8r[] = { -1, -1, -1, 0, 0, 1, 1, 1 };
const int dir_8c[] = { -1, 0, 1, -1, 1, -1, 0, 1 };

int val[MAXN];//堆的价值
bool dp[MAXN];//dp[i]表示是否能填满价值为i的背包

int main() {
	int sum, v, index;

	for (int kase = 1; ; ++kase) {
		sum = index = 0;
		memset(val, 0, sizeof(val));

		for (int i = 1; i <= 6; ++i) {
			scanf("%d", &v);
			sum += v * i;
			//二进制分堆
			int base = 1;
			while (v >= base) {
				val[index++] = base * i;
				v -= base;
				base <<= 1;
			}
			if (v)
				val[index++] = v * i;
		}

		if (!sum)
			break;
		printf("Collection #%d:\n", kase);
		if (sum & 1)
			printf("Can't be divided.\n\n");
		else {
			memset(dp, false, sizeof(dp));
			//解多重背包
			dp[0] = true;
			for (int i = 0; i < index; ++i)
				for (int j = sum; j >= val[i]; --j)
					if (dp[j - val[i]])
						dp[j] = true;

			if (dp[sum / 2])
				printf("Can be divided.\n\n");
			else
				printf("Can't be divided.\n\n");
		}
	}

	//system("pause");
	return 0;
}