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Accepted Necklace

时间:2019-02-13 21:49:55      阅读:176      评论:0      收藏:0      [点我收藏+]

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;
}

 

 

Accepted Necklace

原文:https://www.cnblogs.com/czy-power/p/10371919.html

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