poj2184 Cow Exhibition（背包）

一个背包问题

给出n只奶牛的IQ ,EQ 要求在IQ EQ 都不小于0 的情况下总和最大

dp[i][q] 表示前i只奶牛智商总和为q的最大情商。

对每一只奶牛枚举一下智商的范围， 从-1000*100 ~　1000*100 用数组平移一下 就是0 ~2*1000*100

初始化dp[0]= 0 表示在不装任何奶牛的情况下智商和0，情商和0

要求每次必须正好装满智商

题目：

Cow Exhibition

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8292		Accepted: 3059

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 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.

Source

USACO 2003 Fall

 

代码：

#include <iostream>
using namespace std;

#define MAX 100*2000+5
#define INF 999999999
int dp[MAX];
void init()
{
    for(int i=0;i<MAX;i++)
    {
        dp[i] = -INF;
    }
    dp[100000] = 0;
}
int main()
{
    init();
    int n;
    cin>>n;
    int iq,eq;
    for(int i=0;i<n;i++)
    {
        cin>>iq>>eq;

        if( iq>0)
        {
            for(int e = 2000*100 ; e>=iq;e--)
            {
                if( dp[ e- iq] !=-INF)
                    dp[e] = max( dp[e],dp[e-iq] + eq);
            }
        }
        else
        {
            for(int e = 0;e<=2000*100+iq;e++)
            {
                if( dp[e-iq]!=-INF)
                {
                    dp[e] = max( dp[e] , dp[e-iq]+eq);
                }
            }
        }
    }
    int ans = -1;
    for(int i=100000;i<=200000;i++)
    {
        if(dp[i]>=0)
            ans = max( ans ,  dp[i]+i-100000);
    }
    cout<<ans<<endl;
    return 0;
}