Cow Exhibition_背包（负数情况）

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. 

【题意】给出n头牛，分别给出它们的si，fi；取出几头使得si和fi的和最大，并且si之和与fi之和不能为负数；

【思路】背包题，就是存在负数的情况，可以把负数存入下标，数组开大点。用dp[i]存放每个s[i]能得到的最大的f，根据dp的有无，选出最大的dp

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int N=200005;
const int inf=1<<30;
int dp[N];
int s[202],f[202];
void init()
{
    for(int i=0;i<=200000;i++)
        dp[i]=-inf;
    dp[100000]=0;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        init();
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&s[i],&f[i]);
        }
        for(int i=1;i<=n;i++)
        {
            if(s[i]<0&&f[i]<0) continue;//两个都为负，没有必要进行下去
            if(s[i]>0)//si为正，进行从大到小的背包
            {
                for(int j=200000;j>=s[i];j--)
                {
                    if(dp[j-s[i]]>-inf)
                        dp[j]=max(dp[j],dp[j-s[i]]+f[i]);
                }
            }
            else//为负数则从小到大背包
            {
                for(int j=s[i];j<=200000+s[i];j++)

                {
                    if(dp[j-s[i]]>-inf)
                    dp[j]=max(dp[j],dp[j-s[i]]+f[i]);
                }
            }


        }
        int ans=-inf;
        for(int i=100000;i<=200000;i++)

//dp[]的范围是100000~200000；i就是s[i]，如果此时dp[i]也就是f[i]大于等于0的话，
//再加上s[i]-100000（界限）就是答案

  { 
     if(dp[i]>=0) ans=max(ans,dp[i]+i-100000); } 
     printf("%d
",ans); 
   } return 0; 
}