POJ 2184-cow Exhibition (01 backpack) Problem Solving report

Cow exhibition

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10279		Accepted: 4016

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.

Test instructions

Give you n cows, each cow has a smart value SI and an interesting value fi. Ask you to remove a number of cows from them, so that the si+fi of these cows is the largest and the sum of the SI sum >=0,fi also >=0. The maximum output value.

Exercises

This is the 01 backpack, the idea is to use dp[s] to find out when the smart value of S is the biggest interesting value.

Recursive equation: Dp[s+j]=max (dp[s+j],dp[j]+t).

For ease of handling, we set the origin at 100000, making the s+j in Dp[s+j] always positive.

Reference code:

#include <stdio.h> #define INF 999999999#define pos 100000#define M 200005#define max (b) A>b?a:bint Dp[m];int Main () {int n,s,t,max1,min1,ans;while (~scanf ("%d", &n)) {max1=min1=pos;for (int i=1;i<m;i++)    dp[i]=-inf;        dp[pos]=0;        for (int i=1;i<=n;i++)        {        scanf ("%d%d", &s,&t), if (s>=0) {for (int j=max1;j>=min1;j--) {Dp[j+s ]=max (dp[j+s],dp[j]+t);} Max1+=s;} else {for (int j=min1;j<=max1;j++) {Dp[j+s]=max (dp[j+s],dp[j]+t);} min1+=s;}        }        ans=0;for (int i=pos;i<=max1;i++) {    if (dp[i]>=0)        

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POJ 2184-cow Exhibition (01 backpack) Problem Solving report