HDU 2184 - Cow Exhibition(背包)

最新推荐文章于 2019-07-08 16:12:06 发布

HonniLin

最新推荐文章于 2019-07-08 16:12:06 发布

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分类专栏： DP 文章标签： POJ DP

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本文介绍了一个关于选择一组具有特定智能值和乐趣值的牛参加展览的问题，并将其转化为一种特殊的背包问题进行解决，通过动态规划算法求得总智能值与乐趣值之和的最大值。

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Cow Exhibition

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9217		Accepted: 3507

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

Sample Output

题意：

给出n组（1<=n<=100）数据，有s值（-1000<=s<=1000）和f 值（-1000<=f<=1000) 选择几组，使得s和以及f和都不小于0 且总和最大。

思路：

题目可以转化为背包问题，把s 值作为重量，f值作为价值，因为s值可能为负的，所以将数字前推100000，以dp[100000] 作为原点，

需要注意的是当s[i]大于 0 时，逆推，当 s[i]小于等于0时正推。dp[j] = max (dp[j], dp[i-s[i]]+f[i]);

code:

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
const int inf=0xfffffff;
typedef long long ll;
using namespace std;

int s[105], f[105];
int dp[200005];
int main()
{
    //freopen("in", "r", stdin);
    int n;
    scanf("%d", &n);
    for(int i = 0; i < n; i ++){
        scanf("%d %d", &s[i], &f[i]);
    }
    fill(dp, dp+200005, -inf);
    dp[100000] = 0;
    for(int i = 0; i < n; i ++){
        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 ans = 0;
    for(int 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;
}