Los weight weighing Valley P1441
\ (n-\) range is \ (n-\ Le 20 is \) , \ (m \) in the range of \ (m \ Le. 4 \) .
Violence removing weights traversing each case, a total of \ (n ^ m \) case.
Solving for the remaining number of weights may be combined weight species. Bitset solved using the optimization, the \ (I \) bit \ (1 \) represents a weight \ (I \) may be combined with it. \ (1 \) digit is the final answer.
The initial \ (the bitset [0] =. 1 \) , for the newly added weight \ (I \) , weight \ (A [I] \) , is updated to \ (bitset = bitset \ | \ (bitset << a [ i]) \) .
Time complexity is \ (O (n ^ m \ times n \ times \ text {bitset single operation time}) \) .
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<bitset>
using namespace std;
const int maxn = 25;
const int maxm = 2005;
int n, m, ans, sum;
int vis[maxn], a[maxn], f[maxm];
void solve()
{
bitset<maxm> cnt;
cnt[0] = 1;
for(int i = 1; i <= n; i++){
if(vis[i] == 1) continue;
cnt = cnt | (cnt << a[i]);
}
int ret = cnt.count();
ans = max(ans, ret - 1);
}
void dfs(int now, int step)
{
if(step == m + 1){
solve();
return;
}
for(int i = now; i <= n; i++){
vis[i] = 1;
dfs(i + 1, step + 1);
vis[i] = 0;
}
}
int main()
{
scanf("%d%d", &n, &m);
sum = 0;
for(int i = 1; i <= n; i++){
scanf("%d", &a[i]);
sum += a[i];
}
ans = 0;
dfs(1, 1);
printf("%d\n", ans);
return 0;
}