01背包变形 POJ - 2184 -蒲公英云

01背包变形 POJ - 2184 

野性酷女 2022-06-17 09:26 241阅读 0赞

“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

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.

#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <algorithm>
#define LL long long
#define INF 0x7fffffff
#define MAX 200010
using namespace std;
int dp[MAX];
int s[205];
int f[205];
int n;
int main()
{
    scanf("%d",&n);
    for(int i=1; i<=n; i++)
        scanf("%d%d",&s[i],&f[i]);
    for(int i=0; i<=200000; i++)
        dp[i]=-INF;
              dp[100000]=0;  //以100000为s正负的分界  以f作为价值 找最优解
    for (int i=1; i<=n; i++)
    {
        if(s[i]<0 && f[i]<0) continue;
        if(s[i]>0)
        {
            for(int j=200000; j>=s[i]; j--)
                if(dp[j-s[i]]!= -INF)
                        dp[j]=max(dp[j],dp[j-s[i]]+f[i]);
            }
        else
        {
            for(int j=0; j<=200000+s[i]; j++)
                    if(dp[j-s[i]]!=-INF)
                    dp[j]=max(dp[j],dp[j-s[i]]+f[i]);
        }
    }
    int maxx=-INF;
    for(int i=100000;i<=200000;i++) //在和s的值大于0的情况下，找最大的和f
    {
        if(dp[i]>=0)
        maxx=max(maxx,dp[i]+i-100000);
    }
    printf("%d\n",maxx);
    return 0;
}
/*题意：有n头奶牛，每个奶牛有智慧si和幽默fi这两个属性，
现在需要在这群奶牛里面挑选奶牛参加展览，
要求在满足sum_s和sum_f都大于等于0的情况下，使得sum_s+sum_f的值最大，最大值是多少？*/