Codeforces #380 Subordinates(贪心 构造)

从前往后扫，找到一出现次数为0的数，从后面找一个出现不为0的数转化而来。设置两指针l, r来处理。

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<utility>
using namespace std;
typedef long long LL;
const int N = 1000008, INF = 0x3F3F3F3F;
int a[N], cnt[N], rem[N];
int main(){
    int n, s;
    cin >> n >> s;
    for(int i = 1; i <= n; i++){
        scanf("%d", &a[i]);
    }
    int ans = 0;
    if(a[s]){
        ans++;
        a[s] = 0;
    }
    for(int i = 1; i <= n; i++){
        cnt[a[i]]++;
    }
    cnt[n] += cnt[0] - 1;
    int l = 1, r = n;
    while(l < r){
        while(cnt[l] && l < r){
            l++;
        }
        while(cnt[r] == 0 && l < r){
            r--;
        }
        if(l < r){
            cnt[l]++;
            cnt[r]--;
            ans++;
        }
    }
    cout<<ans<<endl;

    return 0;
}