Subordinates CodeForces

大意: 求构造一棵树, 每个节点回答它的祖先个数, 求最少打错次数.

挺简单的一个构造, 祖先个数等价于节点深度, 所以只需要确定一个最大深度然后贪心即可.

需要特判一下根的深度, 再特判一下只有一个结点的情况

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
//head




#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif


int n, s, f;
int d[N], sum[N], r[N];

int main() {
	scanf("%d%d", &n, &s);
	REP(i,1,n) {
		int t;
		scanf("%d", &t);
		if (i==s&&t) f=1;
		if (i!=s&&!t) ++d[n];
		if (i!=s) ++d[t];
	}
	PER(i,1,n) sum[i]=sum[i+1]+d[i];
	REP(i,1,n) r[i]=r[i-1]+!d[i];
	int ans = n-1;
	REP(i,1,n-1) ans = min(ans, max(sum[i+1],r[i]));
	printf("%d
", ans+f);
}