poj 2184 Cow Exhibition(01背包变形)

2019年04月13日 算法 ⁄ 共 2430字 ⁄ 字号 评论关闭
Cow Exhibition
 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8273 Accepted: 3047

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

```#include<algorithm>
#include<iostream>
#include<string.h>
#include<sstream>
#include<stdio.h>
#include<math.h>
#include<vector>
#include<string>
#include<queue>
#include<set>
#include<map>
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=1e-8;
const double PI=acos(-1.0);
const int maxn=100010;
typedef __int64 ll;
int dp[maxn<<1],s,f;
int main()
{
int i,j,n,l,r,ans;

while(~scanf("%d",&n))
{
l=r=maxn;//l,r为上下界
for(i=1;i<=n;i++)
{
scanf("%d%d",&s[i],&f[i]);
if(s[i]<0)//这样处理比较好
l+=s[i];
else
r+=s[i];
}
memset(dp,0xcf,sizeof dp);//初始化为无穷大
dp[maxn]=0;
for(i=1;i<=n;i++)
{
if(s[i]>0)//考虑无后效性。
{
for(j=r;j>=s[i]+l;j--)
dp[j]=max(dp[j],dp[j-s[i]]+f[i]);
}
else
{
for(j=l+s[i];j<=r;j++)
dp[j]=max(dp[j],dp[j-s[i]]+f[i]);
}
}
ans=0;
for(i=maxn;i<=r;i++)
if(dp[i]>=0)
ans=max(ans,dp[i]+i);
if(ans>=maxn)
ans-=maxn;
printf("%d\n",ans);
}
return 0;
}
```