poj 2184 Cow Exhibition

Cow Exhibition

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5834		Accepted: 1945

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

题意大概是：在所给牛中选若干头牛，在保证fun和smart不小于0的情况下，和最大
思路：其实这道题和普通0-1背包差不多，只是要转换下思想，就是在求fun[j]中可以达到的最大smart。
因为如果取两者和值为状态变量的话，会导致状态不完整，大打个比方，如果f[200]中已经保存了一个值100，当再遇到有值120得到f[200] 时，f[200]的状态就会不完整了。其实和跳舞机差不多，是多阶段的，即由当前fun和smart值决定下一步决策，与之前无关。
另外这里面还有个小陷阱就是当fun为负时，需要递增递推。想想0-1背包的一维状态转移方程是如何得到就知道了

#include <stdio.h>
#define INF (1<<30)
int f[201010];
int main()
{
    int n,i,j,a,b;

    while(scanf("%d",&n)!=EOF)
    {
        for(i=0;i<=201010;i++)
             f[i]=-INF;
           f[100000]=0;

        for(i=0;i<n;i++)
        {
            scanf("%d%d",&a,&b);

            if(a<0&&b<0)continue;

            if(a>0)
            for(j=200000;j>=a;j--)
            {
                if(f[j-a]!=-INF&&f[j-a]+b>f[j])
                   f[j]=f[j-a]+b;
            }
            else
            for(j=a;j<=200000+a;j++)
            {
                if(f[j-a]!=-INF&&f[j-a]+b>f[j])
                   f[j]=f[j-a]+b;
            }

        }
        int max=-INF;
        for(i=100000;i<=200000;i++)
          {
              if(f[i]>=0&&max<f[i]+i-100000)
                     max=f[i]+i-100000;
          }
        printf("%d\n",(max==INF)?0:max);
    }
    return 0;
}