【dfs+dp】砝码称重

 P1441 砝码称重

思路：dfs枚举去掉哪些砝码， 01背包求方案数， 各种情况取max记为ans输出√

边界情况处理不好交了三遍QAQ

dp[j] = dp[j] + dp[j - a[i]] 选上这个砝码的情况+ 不选的情况

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int sz = 2020;
int n, m, ans = 0, sum = 0;
int dp[sz], a[sz];
bool book[sz];
void work(int tot) {
    memset(dp, 0, sizeof(dp));
    int cnt = 0;
    dp[0] = 1;
    for(int i = 1; i <= n; i++) {
        if(book[i]) continue;
        for(int j = tot; j >= 0; j--) {
            if(j >= a[i]) dp[j] = dp[j] + dp[j - a[i]];
        }
    }
    for(int i = 1; i <= tot; i++)
        if(dp[i]) cnt++;
    ans = max(ans, cnt);
}
void dfs(int cur, int now) {//cur当前在第几个砝码前， now放弃了几个砝码 
    if(now > m) return;
    if(cur == n+1) {//是n+1不是n 
        if(now == m) work(sum);
        return;
    }
    dfs(cur+1, now);
    book[cur] = true;//放弃这个砝码
    sum -= a[cur];
    dfs(cur+1, now+1);
    book[cur] = false;
    sum += a[cur];
}
int main() {
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        sum += a[i];
    }
    dfs(1, 0);
    printf("%d", ans);
    return 0;
}