POJ 1014 Dividing(多重背包, 倍增优化)

Q: 倍增优化后, 还是有重复的元素, 怎么办

A: 假定重复的元素比较少, 不用考虑

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.

Sample Input

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

Sample Output

Collection #1:
Can't be divided.

Collection #2:
Can be divided.

思路:

1. 倍增优化, 将 n 转化成 1, 2, 4 ..2^i , (n-前面的和), 然后应用 01背包问题处理

总结:

1. 判断恰好装满的条件为 dp[V] == V. 因为未初始化为 INF, 初始化为 INF 有个好处, 就是可以直接返回 dp[V], 但是更新 dp[v] 时需要加 dp[v] == inf 的判断

代码:

#include <iostream>
using namespace std;
int w[10];
int marble[10000];
int totalWeight;
int dp[120000];
int solve_dp() {
	
	int len = 0;
	for(int i = 1; i <= 6; i ++) {
		int sum = 0;
		for(int j = 0;; j ++) {
			if(sum + (1<<j) > w[i])
				break;
			marble[len++] = (1<<j)*i;
			sum += (1<<j);
		}
		if(sum < w[i])
			marble[len++] = (w[i]-sum)*i;
	}
	memset(dp, 0, totalWeight*sizeof(int));
	// 01 背包
	int V = totalWeight>>1;
	dp[0] = 0;
	for(int i = 0; i < len; i ++) {
		for(int v = V; v >= marble[i]; v--) {
			dp[v] = max(dp[v], dp[v-marble[i]]+marble[i]);
		}
	}
	return (dp[V]==V);
}
int main() {
	freopen("E:\Copy\ACM\测试用例\in.txt", "r", stdin);
	int tc = 0;
	do {
		int sum = 0;
		for(int i = 1; i <= 6; i ++) {
			scanf("%d", &w[i]);
			sum += w[i]*i;
		}
		if(sum == 0)
			return 0;
		tc ++;
		if(sum & 1) { // 为奇数 
			printf("Collection #%d:
Can't be divided.

", tc);
			continue;
		}
		// 重建 model, 转移成 01 背包问题
		totalWeight = sum;
		int ans = solve_dp();
		if(!ans)
			printf("Collection #%d:
Can't be divided.

", tc);
		else
			printf("Collection #%d:
Can be divided.

", tc);
	}while(1);
	return 0;
}