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01背包變形 HDU 2660

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01背包變形 HDU 2660

題目跟一般的01背包相比就是多了個條件。要求最多只能選擇k個，所以只要對著k個進行一次循環就可以了。

Accepted Necklace

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2701 Accepted Submission(s): 1067

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

Sample Output

Source HDU男生專場公開賽——趕在女生之前先過節（From WHU）
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#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define MAXN 22
struct node{
    int val;
    int w;
}a[MAXN];
int dp[1010][22];

int max1(int a,int b){
    return a>b?a:b;
}

int main(){
    int T;
    int n,k,W;
    int i;

    while(~scanf("%d",&T)){
        while(T--){
            scanf("%d%d",&n,&k);
            memset(dp,0,sizeof(dp));
            memset(a,0,sizeof(a));
            for(i=1;i<=n;i++){
                scanf("%d%d",&a[i].val,&a[i].w);
            }
            scanf("%d",&W);
            int j,v;
            for(i=1;i<=n;i++){
                for(v=W;v>=a[i].w;v--){
                    for(j=1;j<=k;j++){
                        dp[v][j] = max1(dp[v][j],dp[v-a[i].w][j-1]+a[i].val);
                    }
                }
            }
            printf("%d\n",dp[W][k]);
        }
    }

    return 0;
}