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 is both smart and fun. In order to does this, Bessie have organized an exhibition that would be put on by the cows. She has given each of the N (1 <= n <=) Cows a thorough interview and determined the values for each cow:the SM Artness si ( -1000 <= si <=) of the cow and the funness fi ( -1000 <= fi <=) of the cow.
Bessie must choose which cows she wants to bring to her exhibition. She believes the total smartness TS of the group are the sum of the Si ' s and, likewise, the total funness TF of the GR OUP 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 n On-negative (since she must also show that the cows is 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 is 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 would improve the value
of TS+TF to ten, but the new value of TF would was negative, so it's not
Allowed. The main idea: there are two sets of data, the two sets of data to be added after the maximum value (must satisfy both sets of data and are positive) a[i] as a cost, b[i] as a value, because there is a negative, so the discussion, if A[i] greater than 0 is equivalent to the normal 01 backpack, if A[i] less than 0, It is the inverse of the 01 backpack (filling space is valuable, that is, to calculate the maximum amount of space to fill the value, while the 01 backpack is the largest consumption of space value) as long as the For loop upside down, 01 backpack full backpack for the loop conditions are in (0 ... V), SO (a[i],a[i]+200000) (separated from the complete backpack, the complete backpack is (0). V) as the following table is not negative so write (A[i] ... V)), so each dp[i] is the maximum value of the subscript, and finally, as long as the traversal, find dp[i]+i-100000 is (if (DP) and I from 100000 traversal to ensure that no sum is negative case) more difficult to think, negative processing as long as the given median value, Backpack size 100000 = 100*1000. (Why?) 01 Backpack Full Backpack for condition must be inside (0,v) interval)
#include <cstdio>#include<cstring>#include<algorithm>using namespacestd;Const intINF =0x3f3f3f3f;Const intMAX =100000+5;intdp[2*MAX];inta[ the],b[ the];intMain () {intx, y; intT; while(~SCANF ("%d",&T)) {memset (A,0,sizeof(a)); memset (b,0,sizeof(b)); for(inti =1; I <= T; i++) {scanf ("%d%d",&a[i],&B[i]); } for(inti =0; I <=200000; i++) Dp[i]= -inf; dp[100000] =0; for(inti =1; I <= T; i++){ if(A[i] <0&& B[i] <0) Continue; Else if(A[i] >0){ for(intj =200000; J >= A[i]; j--){ if(Dp[j-a[i]) >-inf) {Dp[j]= Max (Dp[j],dp[j-a[i]) +B[i]); } } } Else { for(intj = A[i]; J <=200000+a[i]; J + +){ if(Dp[j-a[i]) >-inf) {Dp[j]= Max (dp[j],dp[j-a[i]]+B[i]); } } } } intMAX1 =-inf; for(inti =100000; I <=200000; i++){ if(Dp[i] >=0) Max1= Max (max1,dp[i]+i-100000); } printf ("%d\n", MAX1); } return 0;}View Code
poj2184--Backpack Dp--cow Exhibition
