SuperSale
Description
There is aSuperSalein aSuperHiperMarket. 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 thatSuperHiperMarket. Every person can take as many objects, as he/she can carry out from theSuperSale. 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 atSuperSale?
Input
The input consists ofTtest 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 asingle integer numberNthat indicates the number of objects (1 <= N <= 1000). Then followsNlines, each containing two integers: P and W. The first integer (1<=P<=100) corresponds to the price of object. The secondinteger (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 thisi-thperson 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
2372 1744 2331 24126664 2685 2252 499 1839 1354 9423202026
Sample Output
72514
题意；有n种东西的价值p和重量w,有m个人，给你每个人最多能背的重量，求最大价值。
题解:m重01背包。
#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<cstdlib>#define N 1010#define ll long longusing namespace std;int n,m;int p[N],w[N];int dp[N];int main() {int t;cin>>t;while(t–) {scanf("%d",&n);for(int i=0; i<n; i++)scanf("%d%d",&p[i],&w[i]);scanf("%d",&m);int ans=0;while(m–){int x;scanf("%d",&x);memset(dp,0,sizeof dp);for(int i=0;i<n;i++){for(int j=x;j>=w[i];j–)dp[j]=max(dp[j],dp[j-w[i]]+p[i]);}ans+=dp[x];}cout<<ans<<endl;}}
，收敛自己的脾气，偶尔要刻意沉默，
