POJ 2184 Cow Exhibition

 POJ 2184 Cow Exhibition

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Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6899		Accepted: 2437

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

USACO 2003 Fall

 

题意 ：给出num(num<=100)头奶牛的S和F值(-1000<=S,F<=1000),要求在这几头奶牛中选出若干头，使得在其总S值和总F值均不为负的前提下，求最大的总S值+总F值

 

01背包 可以把Si看成重量 Fi看成价值 处理一下负数  Si<0 正着来 因为每个dp[i]都会受到后面数的影响 Si>0到着来，与前面正好相反 

 

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 const int inf=0x3f3f3f3f;
 7 const int st=100000;  //st必须大于等于100*1000 因为把负数全部容量转化为正数需要 100*1000
 8 int dp[200007];
 9 int a[1009],b[1009];
10 int main()
11 {
12     int p,n;
13     while(cin>>n)
14     {
15         for(int i=0;i<n;i++)
16         cin>>a[i]>>b[i];
17         for(int i=0;i<=200007;i++) dp[i]= -inf;
18         dp[st]=0;
19         for(int i=0;i<n;i++)
20         {    
21             if(a[i]<0&&b[i]<0) continue; 
22             if(a[i]>0)
23             {
24                 for(int j=200000;j>=a[i];j--)
25                 {    
26                     if(dp[j-a[i]]>-inf)  //inf为无限小避免当dp[j-a[i]==无限小时 dp[j]<dp[j-a[i]]+b[i]
27                     dp[j]=max(dp[j],dp[j-a[i]]+b[i]);
28                 }
29                 
30             }
31             else 
32             {
33                 for(int j=0;j<=a[i]+200000;j++)
34                 {    
35                     if(dp[j-a[i]]>-inf)
36                     dp[j]=max(dp[j],dp[j-a[i]]+b[i]);
37                 }
38                 
39             }
40             
41         }
42         int ans=0;
43         for(int i=100000;i<=200000;i++)
44         {
45             if(dp[i]>=0) ans=max(ans,dp[i]+i-100000); 
46         }
47         cout<<ans<<endl;
48         
49     }
50     
51     return 0;
52  } 

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