USACO SECTION 2.2 Subset Sums

最新推荐文章于 2024-03-19 19:37:15 发布

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最新推荐文章于 2024-03-19 19:37:15 发布

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分类专栏： USACO 文章标签： output file integer input each up

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本文讨论了使用动态规划解决子集求和问题的方法，包括从深度优先搜索改进到使用DP表来优化计算过程。通过实例展示了如何计算一组连续整数集合中能被分为两个具有相同和的子集的数量。

Subset Sums
JRM

For many sets of consecutive integers from 1 through N (1 <= N <= 39), one can partition the set into two sets whose sums are identical.

For example, if N=3, one can partition the set {1, 2, 3} in one way so that the sums of both subsets are identical:

{3} and {1,2}

This counts as a single partitioning (i.e., reversing the order counts as the same partitioning and thus does not increase the count of partitions).

If N=7, there are four ways to partition the set {1, 2, 3, ... 7} so that each partition has the same sum:

{1,6,7} and {2,3,4,5}
{2,5,7} and {1,3,4,6}
{3,4,7} and {1,2,5,6}
{1,2,4,7} and {3,5,6}

Given N, your program should print the number of ways a set containing the integers from 1 through N can be partitioned into two sets whose sums are identical. Print 0 if there are no such ways.

Your program must calculate the answer, not look it up from a table.

PROGRAM NAME: subset

INPUT FORMAT

The input file contains a single line with a single integer representing N, as above.

SAMPLE INPUT (file subset.in)

OUTPUT FORMAT

The output file contains a single line with a single integer that tells how many same-sum partitions can be made from the set {1, 2, ..., N}. The output file should contain 0 if there are no ways to make a same-sum partition.

SAMPLE OUTPUT (file subset.out)

一开始的DFS，各种剪枝后还是果断超时

代码如下：

/*
ID: conicoc1
LANG: C
TASK: subset
*/
#include<stdio.h>
#include<stdlib.h>
#include<string.h>

int Number[40];  
int N;
long long int count=0;
int Sum;


void DFS(int k,int SumOfSet)
{
	int i,j;
	int temp;
	if((Sum/2-SumOfSet)<k-1)
		k=Sum/2-SumOfSet+1;
	for(i=k-1;i>=1;i--)
	{	
			SumOfSet+=i;
			
			/*********??????*/
			temp=SumOfSet;
			for(j=i-1;j>=1;j--)
			{
				temp+=j;
				if(temp>=Sum/2)
					break;
			} 
			if(temp<Sum/2)
				break;
			/*****************/
			if(SumOfSet<Sum/2)
				DFS(i,SumOfSet);
			if(SumOfSet==Sum/2)
			{
				count++;
			}
				
			SumOfSet-=i;
	}
}

int main()
{
	FILE *fin,*fout;
	fin=fopen("subset.in","r");
	fout=fopen("subset.out","w");
	memset(Number,0,sizeof(Number));
	
	fscanf(fin,"%d",&N);
	
	int i;
	for(i=1,Sum=0;i<=N;i++)
		Sum+=i;
		
	if(Sum%2==0)
		DFS(N+1,0);
	
	fprintf(fout,"%lld\n",count/2);
	
	return 0;
}

改用DP:

/*
ID: conicoc1
LANG: C
TASK: subset
*/
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
 
int N;
long long int count=0;
int Sum;
long long int A[40][391];  //已决定了N个数，合为M时的情况 


void subset()
{
	int i,j;
	for(i=0;i<=N;i++)
		A[i][0]=1;   //决定i个数，和为0的方案数都只有唯一1种（都不取）
	for(i=1;i<=N;i++)
		A[0][i]=0; 
   /*边界条件*/ 
	
	for(i=1;i<=N;i++)
	{
		for(j=1;j<=Sum;j++)
		{
			if(j-i>=0)
				A[i][j]=A[i-1][j-i]+A[i-1][j];    
			else
			   A[i][j]=A[i-1][j];
		}	
	}
	count=A[N][Sum];
}

int main()
{
	FILE *fin,*fout;
	fin=fopen("subset.in","r");
	fout=fopen("subset.out","w");
 	memset(A,0,sizeof(A));
 	
	fscanf(fin,"%d",&N);
	 
	int i;
	for(i=1,Sum=0;i<=N;i++)
		Sum+=i;
	
	if(Sum%2==0)
	{
		Sum/=2;
		subset();
	}	
  
	fprintf(fout,"%lld\n",count/2);
	
	return 0;	
}

做完这题我觉得得先去NOCOW上面好好理解一下背包问题

动态规划的题目据说千变万化啊= =