Uva--10130（动规）

最新推荐文章于 2020-10-29 11:24:19 发布

转载最新推荐文章于 2020-10-29 11:24:19 发布 · 114 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/naturepengchen/articles/3887084.html

本文介绍了一个经典的01背包问题解决方案，旨在帮助一个家庭在SuperSale活动中最大化所购商品的价值。通过计算不同重量限制下可携带物品的最大价值，最终确定整个家庭能够购买的商品最大价值。

SuperSale

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 Specification

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 Nlines, 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) its 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 Specification

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

72 17

44 23

31 24

64 26

85 22

52 4

99 18

39 13

54 9

Output for the Sample Input

514

思路：经典01背包，先算出所有背包大小的情况，再根据输入逐个相加即可。

 1 /*************************************************************************
 2     > File Name: a.cpp
 3     > Author: Nature
 4     > Mail: 564374850@qq.com 
 5     > Created Time: Wed 30 Jul 2014 12:38:09 PM CST
 6 ************************************************************************/
 7 
 8 #include <cstdio>
 9 #include <cstring>
10 #include <cstdlib>
11 #include <cmath>
12 #include <iostream>
13 #include <algorithm>
14 using namespace std;
15 const int MW = 35;
16 
17 int main(){
18     int v[1005];
19     int w[1005];
20     int dp[30200];
21     int Case;
22     int n,g,sumw;
23     scanf("%d",&Case);
24     while(Case--){
25         scanf("%d",&n);
26         sumw = 0;
27         memset(dp,0,sizeof(dp));
28         for(int i = 0; i < n; ++i)
29             scanf("%d%d",&v[i],&w[i]);
30         for(int i = 0; i < n; ++i){
31             for(int j = MW; j >= w[i]; --j){
32                 dp[j] = max(dp[j],dp[j - w[i]] + v[i]);
33             }
34         }
35         scanf("%d",&g);
36         int ans = 0,tw;
37         while(g--){
38             scanf("%d",&tw);
39             ans += dp[tw];
40         }
41         printf("%d\n",ans);
42     }
43     return 0;
44 }

转载于:https://www.cnblogs.com/naturepengchen/articles/3887084.html