CF755F PolandBall and Gifts

https://www.luogu.com.cn/problem/CF755F

还是有点意思的这题

首先考虑少的，显然每个环先放$c[i]/2$个

然后再把奇环剩下的散点放了

考虑最小的，如果能找到若干个环大小加起来刚好为$k$,那么答案就是$n-k$, 否则就是$n-k-1$

然后这个用二进制分组优化多重背包即可

可以用bitset优化到$n\sqrt{n}logn/w$

#include<bits/stdc++.h>
#define N 1000050
using namespace std;
int n, m, cnt[N], vis[N], a[N], c[N], to[N];
void solve() {
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i ++) vis[i] = cnt[i] = c[i] = a[i] = 0;
    for(int i = 1; i <= n; i ++) scanf("%d", &to[i]);
    int gs = 0;
    for(int i = 1; i <= n; i ++) if(!vis[i]) {
        int x = i;
        c[++ gs] = 0;
        while(!vis[x]) {
            c[gs] ++;
            vis[x] = 1;
            x = to[x];
        }
        cnt[c[gs]] ++;
    }    

    int ans = 0;
    for(int i = 1; i <= gs; i ++) ans += (c[i] >> 1);

    int tot = 0;
    for(int i = 1; i <= n; i ++) if(cnt[i]) {
        for(int j = 1; cnt[i] >= j; j <<= 1)
            a[++ tot] = j * i, cnt[i] -= j;
        if(cnt[i]) a[++ tot] = cnt[i] * i;
    }

    bitset<N> f; f.reset();
    f[0] = 1;
    for(int i = 1; i <= tot; i ++) {
        f |= (f << a[i]);
    }

    if(f[m]) printf("%d ", m);
    else printf("%d ", m + 1);

    if(ans >= m) printf("%d", m << 1);
    else printf("%d ", (ans << 1) + min(m - ans, n - (ans << 1)));
}
int main() {
    solve();
    return 0;
}