Codeforces 755 F. PolandBall and Gifts 多重背包+贪心

F. PolandBall and Gifts

 

It's Christmas time! PolandBall and his friends will be giving themselves gifts. There are n Balls overall. Each Ball has someone for whom he should bring a present according to some permutation p, pi ≠ i for all i.

Unfortunately, Balls are quite clumsy. We know earlier that exactly k of them will forget to bring their gift. A Ball number i will get his present if the following two constraints will hold:

1. Ball number i will bring the present he should give.
2. Ball x such that px = i will bring his present.

What is minimum and maximum possible number of kids who will not get their present if exactly k Balls will forget theirs?

Input

The first line of input contains two integers n and k (2 ≤ n ≤ 106, 0 ≤ k ≤ n), representing the number of Balls and the number of Balls who will forget to bring their presents.

The second line contains the permutation p of integers from 1 to n, where pi is the index of Ball who should get a gift from the i-th Ball. For all i, pi ≠ i holds.

Output

You should output two values — minimum and maximum possible number of Balls who will not get their presents, in that order.

Examples

input

5 2
3 4 1 5 2

output

2 4

Note

In the first sample, if the third and the first balls will forget to bring their presents, they will be th only balls not getting a present. Thus the minimum answer is 2. However, if the first ans the second balls will forget to bring their presents, then only the fifth ball will get a present. So, the maximum answer is 4.

 题意：

　　n个人，每个人固定给一个人送一个gift，不会给自己送gift，不会有一个人收到两个gift

　　也就是说，每个人会收到一个gift和送出去一个gift，这样的人ans+1

　　现在你可以指定k个人不会送出去gift

　　ans最小和最大是多少

题解：

　　划分为多个置换群

　　ans最大去贪心就好了

　　最小的话： 如果选取置换群大小相加存在等于k的最小就是k，否则是k+1

　　利用以上：跑背包，是否能构成K，范围太大，需要二进制优化多重背包优化到n*∑log(b[i])，另外bitset优化 

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const int N = 1e6+10, maxn = 1e3+20, mod = 1e9+7, inf = 2e9;

int a[N],b[N],n,k,vis[N],H[N],cnt;
int main() {
    scanf("%d%d",&n,&k);
    for(int i = 1; i <= n; ++i) scanf("%d",&a[i]);
    for(int i = 1; i <= n; ++i) {
        if(vis[i]) continue;
        int j = a[i],cnts = 0;
        while(!vis[j]) {
            cnts++;
            vis[j] = 1;
            j = a[j];
        }
        b[++cnt] = cnts;
    }
    sort(b+1,b+cnt+1);
    int ans2 = 0,kk = k;
    for(int i = 1; i <= cnt; ++i) {
        if(kk >= b[i]/2) kk-=b[i]/2,ans2 += b[i]/2*2;
        else {
            ans2 += 2*kk;
            kk = 0;
            break;
        }
    }
    ans2 += kk;
    ans2 = min(n,ans2);
    for(int i = 1; i <= cnt; ++i) H[b[i]]++;
    bitset<N> dp;
    dp.set(0);
    for(int i = 1; i <= n; ++i) {
        if(H[i] == 0) continue;
        int t = H[i],l = 1;
        while(t) {
            int w = min(t,l);
            t -= w;
            dp |= dp<<(w*i);
            l <<= 1;
        }
        if(dp[k] == 1) {
            printf("%d %d
",k,ans2);
            return 0;
        }
    }
    printf("%d %d
",1+k,ans2);
    return 0;
}