Accepted Necklace

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4658    Accepted Submission(s): 1853

Problem Description I have N precious stones, and plan to use K of them to make a necklace for my mother, but she won't accept a necklace which is too heavy. Given the value and the weight of each precious stone, please help me find out the most valuable necklace my mother will accept.  

 

Input The first line of input is the number of cases.
For each case, the first line contains two integers N (N <= 20), the total number of stones, and K (K <= N), the exact number of stones to make a necklace.
Then N lines follow, each containing two integers: a (a<=1000), representing the value of each precious stone, and b (b<=1000), its weight.
The last line of each case contains an integer W, the maximum weight my mother will accept, W <= 1000.  

 

Output For each case, output the highest possible value of the necklace.  

 

Sample Input 1 2 1 1 1 1 1 3  

 

Sample Output 1   思路：01背包，另外要开二维数组分别记录当前重量和数量（有数量限制）。

#include<bits/stdc++.h>
#include<queue>
#include<cstdio>
#include<iostream>
#define REP(i, a, b) for(int i = (a); i <= (b); ++ i)
#define REP(j, a, b) for(int j = (a); j <= (b); ++ j)
#define PER(i, a, b) for(int i = (a); i >= (b); -- i)
using namespace std;
template <class T>
inline void rd(T &ret){
    char c;
    ret = 0;
    while ((c = getchar()) < '0' || c > '9');
    while (c >= '0' && c <= '9'){
        ret = ret * 10 + (c - '0'), c = getchar();
    }
}
int dp[1010][30];
struct node{int val,w;}p[30];
int main()
{
    int T;
    rd(T);
    while(T--){
        int n,k,mw;
        rd(n),rd(k);
        for(int i=1;i<=n;i++)cin>>p[i].val>>p[i].w;
        cin>>mw;
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++){
            for(int j=mw;j>=p[i].w;j--){
                for(int h=1;h<=k;h++){
                    dp[j][h]=max(dp[j][h],dp[j-p[i].w][h-1]+p[i].val);
                }
            }
        }
        int maxn=-0xfffffff;
        for(int i=0;i<mw;i++)maxn=max(maxn,dp[i][k]);
        cout<<maxn<<endl;
    }
    return 0;
}

 

 

标签：case,Accepted,necklace,each,Necklace,line,include,my
来源： https://www.cnblogs.com/czy-power/p/10371919.html