Cow Exhibition POJ - 2184

最新推荐文章于 2024-06-20 15:36:36 发布

weixin_30593261

最新推荐文章于 2024-06-20 15:36:36 发布

阅读量194

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/zgglj-com/p/7466474.html

解决一个关于选择一组具有最大综合智能与乐趣值的牛群的问题，同时确保这两项总值均为非负数。通过调整01背包问题的解决方案，实现了一个算法来找到最优解。

"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

Sample Output

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.

orz，觉得是01背包，也想过把负值加一个大数变为正的，然而。。。还是不知道怎么写！！！！

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<cstring>
 5 using namespace std;
 6 
 7 const int INF=0x3f3f3f3f;
 8 const int maxn=105;
 9 
10 int n;
11 int dp[200010],va[maxn],vo[maxn];
12 
13 void solve(){
14     dp[100000]=0;
15     for(int i=0;i<n;i++){
16         if(va[i]>0){
17             for(int j=200000;j>=va[i];j--)
18                 if(dp[j-va[i]]>-INF)
19                     dp[j]=max(dp[j],dp[j-va[i]]+vo[i]);
20         }
21         else{
22             for(int j=0;j<=200000+va[i];j++)
23                 if(dp[j-va[i]]>-INF)
24                     dp[j]=max(dp[j],dp[j-va[i]]+vo[i]);
25         }
26     }
27     
28     int ans=0;
29     for(int i=100000;i<=200000;i++)
30         if(dp[i]>=0&&dp[i]+i-100000>ans)
31             ans=dp[i]+i-100000;
32     cout<<ans<<endl;
33 }
34 
35 int main()
36 {   cin>>n;
37     for(int i=0;i<n;i++) cin>>va[i]>>vo[i];
38     for(int i=0;i<=200000;i++) dp[i]=-INF;
39     solve();
40     
41     return 0;
42 }