首页 > ACM题库 > HDU-杭电 > HDU 1171 Big Event in HDU-背包问题-[解题报告] C++
2013
12-03

HDU 1171 Big Event in HDU-背包问题-[解题报告] C++

Big Event in HDU

问题描述 :

Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don’t know that Computer College had ever been split into Computer College and Software College in 2002.
The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. All facilities must go halves. First, all facilities are assessed, and two facilities are thought to be same if they have the same value. It is assumed that there is N (0<N<1000) kinds of facilities (different value, different kinds).

输入:

Input contains multiple test cases. Each test case starts with a number N (0 < N <= 50 — the total number of different facilities). The next N lines contain an integer V (0<V<=50 –value of facility) and an integer M (0<M<=100 –corresponding number of the facilities) each. You can assume that all V are different.
A test case starting with a negative integer terminates input and this test case is not to be processed.

输出:

For each case, print one line containing two integers A and B which denote the value of Computer College and Software College will get respectively. A and B should be as equal as possible. At the same time, you should guarantee that A is not less than B.

样例输入:

2
10 1
20 1
3
10 1 
20 2
30 1
-1

样例输出:

20 10
40 40

这一道题是一道多重背包,和 hdu1059 是同一个类型的

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define CLR(arr,val) memset(arr,val,sizeof(arr))
const int MAX = 250100;
bool dp[MAX];
int V;

void zeroOnePack(int cost ,int value)
{
	for(int i = V; i >= cost ;i--)
		if(dp[i - cost] == true)
			dp[i] = true;
}
void completePack(int cost,int value)
{
	for(int i = cost ;i <= V ;i++)
		if(dp[i - cost] == true)
			dp[i] = true;
}
void multipPack(int cost ,int value ,int amount)
{
	if(cost * amount >= V)
		completePack(cost,value);
	else
	{
		int k = 1;
		while(k < amount )
		{
			zeroOnePack(cost * k ,value * k);
			amount -= k;
			k *= 2;
		}
		zeroOnePack(cost * amount ,value * amount );
	}
}
int main()
{
	//freopen("in.txt","r",stdin);
	int group_num;
	while(cin>>group_num,group_num > 0)
	{
		CLR(dp,false);
		dp[0] = true;
		int v[510],num[510];
		V = 0;
		for(int i = 1;i <= group_num ;i++)
		{
			scanf("%d%d",&v[i],&num[i]);
			V += (v[i] * num[i]);
		}
		int tmp = V;
		V /= 2; 
		for(int i = 1;i <= group_num ;i++)
		{
			multipPack(v[i],1,num[i]);
		}
		int cur=0;
		for(int i = V;i >= 1 ;i--)
			if(dp[i] == true)
			{
				cur = i;
				break;
			}
		printf("%d %d\n",tmp-cur,cur);
	}
	return 0;
}


  1. 题本身没错,但是HDOJ放题目的时候,前面有个题目解释了什么是XXX定律。
    这里直接放了这个题目,肯定没几个人明白是干啥

  2. 这道题目的核心一句话是:取还是不取。
    如果当前取,则index+1作为参数。如果当前不取,则任用index作为参数。

  3. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。