[POJ 2184] Cow Exhibition

Cow Exhibition

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9479		Accepted: 3653

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

有点不好理解、- -

由于SWUST OJ 数据太水，因而乱写AC、受不了。 - -

还有一种做法就是将每个聪明指数+1000、然后、、受不了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f

int n;
int w[110];
int v[110];
int dp[200010];

int main()
{
    int i,j,k=100000;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d%d",&w[i],&v[i]);
        }
        for(i=0;i<=200000;i++) dp[i]=-INF;
        dp[k]=0;
        for(i=1;i<=n;i++)
        {
            if(w[i]>=0)      //正的是逆序
            {
                for(j=200000;j>=w[i];j--)
                  if(dp[j-w[i]]>-INF)
                    dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
            }
            else             //负的是顺序
            {
                for(j=0;j<=200000+w[i];j++)
                  if(dp[j-w[i]]>-INF)
                    dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
            }
        }
        int ans=0;
        for(i=100000;i<=200000;i++)
            if(dp[i]>=0 && dp[i]+i-100000>ans) ans=dp[i]+i-100000;
        printf("%d\n",ans);
    }
    return 0;
}