【多重背包】二进制拆分【模板】

把多重背包用二进制拆分,拆分后的能表示它所能表示的任意数字. 
比如:7 = 1 + 2 + 4, 13 = 1 + 4 + 8; 
然后把价值和空间对应也更新,然后用01背包直接写就行,复杂度:n∗m∗log(num)n∗m∗log(num).

#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;

typedef long long LL;
const int MAXN = 1e6 + 10;
LL v[110], w[110], c[110];
//多重背包:v表示价值,w表示空间,c表示个数
LL val[1010], size[1010];
//01背包:val表示新的价值,size表示新的空间 
LL dp[MAXN];

int main() { 
    int n, m;
    while(scanf("%d %d", &n, &m) != EOF) {     
        int count = 0;
        for(int i = 0; i < n; i++) {  
            scanf("%lld %lld %lld", &c[i], &w[i], &v[i]);
            for(LL j = 1; j <= c[i]; j << 1) {
                val[count] = j * v[i];
                size[count++] = j * w[i];
                c[i] -= j;
            }
            if(c[i] > 0) {
                val[count] = c[i] * v[i];
                size[count++] = c[i] * w[i];
            }
        }
        memset(dp, 0, sizeof(dp));
        for(int i = 0; i < count; ++i) {
            for(int j = m; j >= size[i]; --j) {
                dp[j] = max(dp[j], dp[j - size[i]] + val[i]);
            }
        }
        printf("%lld\n", dp[m]);
    }  
    return 0;  
}

转自:https://blog.csdn.net/qq_36368339/article/details/81186667

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转载自blog.csdn.net/tingtingyuan/article/details/81539659