poj2184 Cow Exhibition（p-01背包的灵活运用）

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题目链接：

id=2184">http://poj.org/problem?id=2184

Description

"Fat and docile, big and dumb, they look so stupid, they aren't much 

fun..." 

- Cows with Guns by Dana Lyons 



The cows want to prove to the public that they are both smart and fun. In order to do this, Bessie has organized an exhibition that will be put on by the cows. She has given each of the N (1 <= N <= 100) cows a thorough interview and determined two values for
each cow: the smartness Si (-1000 <= Si <= 1000) of the cow and the funness Fi (-1000 <= Fi <= 1000) of the cow. 



Bessie must choose which cows she wants to bring to her exhibition. She believes that the total smartness TS of the group is the sum of the Si's and, likewise, the total funness TF of the group is the sum of the Fi's. Bessie wants to maximize the sum of TS
and TF, but she also wants both of these values to be non-negative (since she must also show that the cows are well-rounded; a negative TS or TF would ruin this). Help Bessie maximize the sum of TS and TF without letting either of these values become negative. 

Input

* Line 1: A single integer N, the number of cows 



* Lines 2..N+1: Two space-separated integers Si and Fi, respectively the smartness and funness for each cow. 

Output

* Line 1: One integer: the optimal sum of TS and TF such that both TS and TF are non-negative. If no subset of the cows has non-negative TS and non- negative TF, print 0. 

Sample Input

5
-5 7
8 -6
6 -3
2 1
-8 -5

Sample Output

8

Hint

OUTPUT DETAILS: 



Bessie chooses cows 1, 3, and 4, giving values of TS = -5+6+2 = 3 and TF 

= 7-3+1 = 5, so 3+5 = 8. Note that adding cow 2 would improve the value 

of TS+TF to 10, but the new value of TF would be negative, so it is not 

allowed. 

Source

field=source&key=USACO+2003+Fall" style="text-decoration:none">USACO 2003 Fall

题意：要求从N头牛中选择若干头牛去參加比赛，如果这若干头牛的智商之和为sumS,幽默度之和为sumF现要求在全部选择中。在使得sumS>=0&&sumF>=0的基础上。使得sumS+sumF最大并输出其值.

思路：事实上这道题和普通0-1背包几乎相同，仅仅是要转换下思想。就是在求fun[j]中能够达到的最大smart。

我们设智商属性为费用。幽默感属性为价值，问题转换为求费用大于和等于0时的费用、价值总和。

代码+解释例如以下：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdlib>
#include <climits>
#include <ctype.h>
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;
#define PI acos(-1.0)
#define INF 0x3fffffff
int MAX(int a,int b)
{
	if( a > b)
		return a;
	return b;
}
int main()
{
	int s[147],f[147],dp[200047];
	int N,i,j;
	while(~scanf("%d",&N))
	{
		for(i = 0 ; i <= 200000 ; i++)
		{
			dp[i] =-INF;
		}
		for(i = 1 ; i <= N ; i++)
		{
			scanf("%d%d",&s[i],&f[i]);
		}
		dp[100000] = 0;//初始化dp[100000]相当于“dp[0]”也就是智力和为零的时候
		for(i = 1 ; i <= N ; i++)
		{
			if(s[i] < 0 && f[i] < 0)
				continue;
			if(s[i] > 0)//假设s[i]为正数。那么我们就从大的往小的方向进行背包
			{
				for(j = 200000 ; j >= s[i] ; j--)//100000以上是智力和为正的时候
				{
					if(dp[j-s[i]] > -INF)
					{
						dp[j]=MAX(dp[j],dp[j-s[i]]+f[i]);
					}
				}
			}
			else//假设s[i]为负数。那么我们就从小的往大的方向进行背包
			{
				for(j = s[i] ; j <= 200000+s[i] ; j++)//100000一下是当智力和为负的时候
				{
					if(dp[j-s[i]] > -INF)
					{
						dp[j]=MAX(dp[j],dp[j-s[i]]+f[i]);
					}
				}
			}
		}
		int ans=-INF;
		for(i = 100000 ; i <= 200000 ; i++)//仅仅在智力和为正的区间查找智力和幽默值的和最大的
		{
			if(dp[i]>=0)
			{
				ans=MAX(ans,dp[i]+i-100000);
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}