Cow Exhibition（DP）

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.

一堆牛，有智商和情商，选出一些牛，在智商和和情商和都不为负数的前提下，求情商和智商和的最大。

昨天写了那么多题，头昏脑涨，今早看了一下发现就是背包，随便以智商或是情商为背包，求在这个智商下，情商的最大值，由于是要把容量完全装满，所以初始化除了dp[0] = 0;其他全为-inf。
然后wa了。
发现负数无法处理，可以看到数据范围是-1e5~1e5，可以加一个偏移量1e5即可。

还有一个优化就是每个情商是-1000到1000的，所以每次扩大搜索范围1000就行了，减少了很多不必要的更新。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<utility>
#include<sstream>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 1005;
const int lea = 1e5;
int dp[200005];
struct Node
{
    int x,y;
}node[105];
int main()
{
    #ifdef LOCAL
    freopen("C:\\Users\\ΡΡ\\Desktop\\in.txt","r",stdin);
    //freopen("C:\\Users\\ΡΡ\\Desktop\\out.txt","w",stdout);
    #endif // LOCAL
    int n,len = 200000;
    scanf("%d",&n);
    for(int i = 1;i <= n;i++)
    {
        scanf("%d%d",&node[i].x,&node[i].y);
    }
    for(int i = 0;i <= len;i++)dp[i] = -inf;
    dp[lea] = 0;
    for(int i = 1;i <= n;i++)
    {
        //if(node[i].x + node[i].y < 0)continue;
        if(node[i].x > 0)
            for(int j = lea + (i - 1)*1000 + node[i].x;j >= lea - (i - 1)*1000 + node[i].x;j--)
                dp[j] = max(dp[j],dp[j - node[i].x] + node[i].y);
        else
        {
            for(int j = lea - (i - 1)*1000 + node[i].x;j <= lea + (i - 1)*1000 + node[i].x;j++)
                dp[j] = max(dp[j],dp[j - node[i].x] + node[i].y);
        }
    }
    int ans = 0;
    for(int i = lea;i <= len;i++)
        if(dp[i] >= 0)
        ans = max(ans,i - lea + dp[i]);
    if(ans >= 0)
        printf("%d\n",ans);
    else
        printf("0\n");
    return 0;
}