UVA 10130 （多个独立01背包）（2015排位赛2、E）

最新推荐文章于 2017-10-30 22:51:40 发布

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本文介绍了一种典型的背包问题，并通过动态规划的方法求解。问题背景设定在一个超级市场的大促销中，每个家庭成员能携带一定重量的商品，目标是最大化商品总价值。文章详细解释了输入输出格式，并给出了一段C++代码实现。

题目：

Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

Submit Status Practice UVA 10130

Description

There is a SuperSale in a SuperHiperMarket. Every person can take only one object of each kind, i.e. one TV, one carrot, but for extra low price. We are going with a whole family to that SuperHiperMarket. Every person can take as many objects, as he/she can carry out from the SuperSale. We have given list of objects with prices and their weight. We also know, what is the maximum weight that every person can stand. What is the maximal value of objects we can buy at SuperSale?

Input

The input consists of T test cases. The number of them (1<=T<=1000) is given on the first line of the input file.

Each test case begins with a line containing a single integer number N that indicates the number of objects (1 <= N <= 1000). Then follows N lines, each containing two integers: P and W. The first integer (1<=P<=100) corresponds to the price of object. The second integer (1<=W<=30) corresponds to the weight of object. Next line contains one integer (1<=G<=100) it’s the number of people in our group. Next G lines contains maximal weight (1<=MW<=30) that can stand this i-th person from our family (1<=i<=G).

Output

For every test case your program has to determine one integer. Print out the maximal value of goods which we can buy with that family.

Sample Input

Sample Output

72
514

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一看到题目描述，便想到是DP。

一开始没看清题目，“Every person can take only one object of each kind”这句话意思是每个人能在自身负载内从每一种物品带走一个。

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代码：

#include
  
   
#include
   
    
#include
    
     
using namespace std;

const int INF = 1000000000;
int P[1000 + 10];
int W[1000 + 10];
int dp[1000 + 10][35];

int main()
{
    int T;
    scanf("%d", &T);
    while( T-- )
    {
        int n;
        scanf("%d", &n);
        for(int i = 0; i < n; i++) {
                scanf("%d%d", &P[i], &W[i]);
        }
        memset(dp, 0, sizeof(dp));
        for(int i = 0; i < n; i++) {
            for(int j = 0; j <= 30; j++) {
                if(j < W[i]) dp[i+1][j] = dp[i][j];
                else dp[i+1][j] = max(dp[i][j], dp[i][j - W[i]] + P[i]);
            }
        }
        int num;
        scanf("%d", &num);
        int ans = 0;
        for(int i = 0; i < num; i++) {
            int k;
            scanf("%d", &k);
            int res = 0;
            for(int j = 0; j <= k; j++) {
                if(res < dp[n][j]) res = dp[n][j];
            }
            ans += res;
        }
        printf("%d\n", ans);
    }
    return 0;
}