P1441 砝碼稱重

思路：dfs列舉去掉哪些砝碼， 01揹包求方案數， 各種情況取max記為ans輸出√

邊界情況處理不好交了三遍QAQ

dp[j] = dp[j] + dp[j - a[i]] 選上這個砝碼的情況+ 不選的情況

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 using namespace std;
 5 const int sz = 2020;
 6 int n, m, ans = 0, sum = 0;
 7 int dp[sz], a[sz];
 
 8 bool book[sz];
 9 void work(int tot) {
10     memset(dp, 0, sizeof(dp));
11     int cnt = 0;
12     dp[0] = 1;
13     for(int i = 1; i <= n; i++) {
14         if(book[i]) continue;
15         for(int j = tot; j >= 0; j--) {
16             if(j >= a[i]) dp[j] = dp[j] + dp[j - a[i]];
17
 
         }
18     }
19     for(int i = 1; i <= tot; i++)
20         if(dp[i]) cnt++;
21     ans = max(ans, cnt);
22 }
23 void dfs(int cur, int now) {//cur當前在第幾個砝碼前， now放棄了幾個砝碼 
24     if(now > m) return;
25     if(cur == n+1) {//是n+1不是n 
26         if(now == m) work(sum);
27         return
 
;
28     }
29     dfs(cur+1, now);
30     book[cur] = true;//放棄這個砝碼
31     sum -= a[cur];
32     dfs(cur+1, now+1);
33     book[cur] = false;
34     sum += a[cur];
35 }
36 int main() {
37     scanf("%d%d", &n, &m);
38     for(int i = 1; i <= n; i++) {
39         scanf("%d", &a[i]);
40         sum += a[i];
41     }
42     dfs(1, 0);
43     printf("%d", ans);
44     return 0;
45 }