POJ--2184 -- Cow Exhibition -- 01 backpack

POJ--2184 -- Cow Exhibition -- 01 backpack

Cow Exhibition

Time Limit:1000 MS		Memory Limit:65536 K
Total Submissions:9429		Accepted:3624

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 wowould 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 78 -66 -32 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 wocould improve the value
Of TS + TF to 10, but the new value of TF wocould be negative, so it is not
Allowed.

Forgive me for my country of love, so I can't calm down and look at the questions, so I will directly find the questions in the Baidu conclusion report on the Internet. The online questions are basically vague, Just Like farting.

S and F are two attributes of the ox, respectively. They tell you the two attributes of the ox, and then ask you to select a random number of cows so that the S and F values of all cows are the largest, at the same time, the S value of all cattle cannot be less than zero, and the F value cannot be less than zero.

Resolution: if it is used as a 01 backpack, one of the attributes will be treated as capacity. In this way, the maximum value will be obtained under a certain capacity.

For example, if S is used as the capacity, the meaning of dp [X] is: When S is set to X, the maximum number of fvalues is reached.

However, if the S value is smaller than zero, we should consider how small the minimum value is!

# Include
  
   
# Include
   
    
# Include
    
     
# Include using namespace std; int dp [222222]; // if the minimum value is-10 W, it is better to leave the entire array with a space of more than 10 W to a negative number. int main (void) {int n, m, I, j, k, l, a, B, x = 111111; // x indicates the position int Max, Min of 0; // The maximum and minimum boundary, it is convenient to traverse the while (scanf ("% d", & n) data in the range that has been evaluated )! = EOF) {for (I = 0; I <222222; I ++) {dp [I] =-999999;} Max = Min = x; dp [x] = 0; for (I = 0; I
     
      
= Min; j --) // positive value: from big to small. 01 backpack {if (dp [j + a]
      
        = 0 & l