HDU-1059 Dividing (多重背包 二進制優化+剪枝)

Dividing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 27707    Accepted Submission(s): 7961

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.

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.

Output For each colletcion, 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.
        

#include <bits/stdc++.h>
using namespace std;
int dp[120005], num[7];
int main(){
	int casenum = 1;
	while(scanf("%d %d %d %d %d %d", &num[1], &num[2], &num[3], &num[4], &num[5], &num[6]) != EOF){
		if(num[1] == 0 && num[2] == 0 && num[3] == 0 && num[4] == 0 && num[5] == 0 && num[6] == 0) return 0;
		int tot = 0;
		for(int i = 1; i <= 6; ++i){
			tot += i * num[i];
		}
		if(tot % 2 == 1){
			printf("Collection #%d:\nCan't be divided.\n\n", casenum++);  
            continue;  
		}
		tot /= 2;
		memset(dp, 0, (tot + 1) * sizeof(dp[0]));
		int tot1 = 0;
		for(int i = 1; i <= 6; ++i){
			tot1 += num[i] * i;
			int cnt = 0, cur = 1, v;
			while(cnt + cur <= num[i]){
				v = cur * i;
				for(int j = min(tot, tot1); j >= v; --j){
					if(dp[j] == j){
						break;
					}
					dp[j] = max(dp[j], dp[j - v] + v);
				}
				cnt += cur;
				cur *= 2;
			}
			if(num[i] > cnt){
				v = i * (num[i] - cnt);
				for(int j = min(tot, tot1); j >= v; --j){
					if(dp[j] == j){
						break;
					}
					dp[j] = max(dp[j], dp[j - v] + v);
				}
			}
		}
		if(dp[tot] == tot){
			printf("Collection #%d:\nCan be divided.\n\n", casenum++); 
		}
		else{
			printf("Collection #%d:\nCan't be divided.\n\n", casenum++);
		}
	}
}

/*
題意：
1.2.3.4.5.6的硬币各不超過20000個，問是否可以分為相等的兩份。

思路：
和0/1背包裡面的分硬币題差不多，隻不過這裡分組背包需要用二進制優化一下複雜度，同時剪枝一下，
dp[j] == j時可以停止更新了，因為是從硬币1到6更新的，小的硬币都可以填滿j了，那麼更小的一定可以填滿。
*/