Cow Exhibition （背包中的负数问题）

个人心得：背包，动态规划真的是有点模糊不清，太过于抽象，为什么有些是从后面递推，

有些状态就是从前面往后面，真叫人头大。

这一题因为涉及到负数，所以网上大神们就把开始位置从10000开始，这样子就转变为了由一个正数背包装的最大值构成的背包问题了，

只要对于正数背包中的容积加价值相加就可以了，还是不太了解，有点模糊不清楚的感觉；

这题还要注意当背包容积大于0小于0的时候的递推，大于0是从后面往前面走，小于0反之。

大于0就是纯粹的01背包问题，和模板一样的吧，小于0的话就好像多重背包吧，可以覆盖，所以就从前面开始递推吧，脑阔痛。

"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. 

 1 #include
 2 #include
 3 #include
 4 #include
 5 #include<string>
 6 #include
 7 using namespace std;
 8 const int money=1000;
 9 const int inf=999999999;
10 int cows[105],cowa[105];
11 int dp[200005];
12 void init(){
13          for(int i=0;i<=200005;i++)
14             dp[i]=-inf;
15             dp[100000]=0;
16 }
17 int main(){
18     int t;
19     while(cin>>t){
20     init();
21     for(int i=1;i<=t;i++)
22         {
23             cin>>cows[i]>>cowa[i];
24 
25         }
26         int maxn=-inf;
27     for(int i=1;i<=t;i++)
28       {
29           if(cows[i]<0&&cowa[i]<0)
30             continue;
31           if(cows[i]>0){
32             for(int j=200000;j>=cows[i];j--)
33                 if(dp[j-cows[i]]>-inf)
34                    dp[j]=max(dp[j],dp[j-cows[i]]+cowa[i]);
35           }
36           else
37           {
38               for(int j=0;j<=200000+cows[i];j++)
39                  if(dp[j-cows[i]]>-inf)
40                    dp[j]=max(dp[j],dp[j-cows[i]]+cowa[i]);
41           }
42 
43       }
44       for(int i=100000;i<=200000;i++)
45         if(dp[i]>=0) maxn=max(maxn,dp[i]+i-100000);
46      cout<endl;
47     }
48     return 0;
49 }

 

转载于:https://www.cnblogs.com/blvt/p/7382458.html