codeforces 1283 C 贪心

贪心瞎搞

题意：给了n个数，对于任意i，ai都可以变成ai+1，ai-1，ai三个位置。问进行任意次操作，最少和最多可以覆盖多少个数。

思路：排序a从小到大，对于每一个ai，优先选择左面的（ai-1）没有选择的位置，最后被标记的位置数列即使最多的个数。

　　优先选择最右面的（ai+1）选择了的位置，如果都没选择，则选择ai+1，然后统计覆盖的数列寄给最少的个数。

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector> 
// #include <bits/stdc++.h>
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define sp ' '
#define endl '\n'
#define inf  0x3f3f3f3f;
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
#define P pair<int, int>
#define fi first
#define se second
#define pb(x) push_back(x)
#define ppb() pop_back()
#define mp(a,b) make_pair(a,b)
#define ms(v,x) memset(v,x,sizeof(v))
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define repd(i,a,b) for(int i=a;i>=b;i--)
#define sca3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define sca2(a,b) scanf("%d %d",&(a),&(b))
#define sca(a) scanf("%d",&(a));
#define sca3ll(a,b,c) scanf("%lld %lld %lld",&(a),&(b),&(c))
#define sca2ll(a,b) scanf("%lld %lld",&(a),&(b))
#define scall(a) scanf("%lld",&(a));


using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a, ll b, ll mod){ll sum = 1;while (b) {if (b & 1) {sum = (sum * a) % mod;b--;}b /= 2;a = a * a % mod;}return sum;}

const double Pi = acos(-1.0);
const double epsilon = Pi/180.0;
const int maxn = 2e5+10;
int n;
int a[maxn];
int vis[maxn];
int d[3] = {-1,0,1};
int main()
{
    //freopen("input.txt", "r", stdin);
    cin>>n;
    rep(i,1,n) cin>>a[i];
    sort(a+1,a+1+n);
    queue<pair<int,int> >que;
    rep(i,1,n){
        rep(j,0,2){
            int now = a[i]+d[j];
            if(vis[now] == 0){
                vis[now]=1;
                break;
            }
        } 
    }
    int ansmax = 0;
    rep(i,0,n+1){
        if(vis[i] == 1){
            ansmax++;
        }
    }
    memset(vis,0,sizeof(vis));
    rep(i,1,n){
        int flag = 0;
        repd(j,2,0){
            int now = a[i]+d[j];
            if(vis[now] == 1){
                flag = 1;
                break;
            }
        } 
        if(flag == 0){
            vis[a[i]+1]=1;
        }
    }
    int ansmin  = 0;
    rep(i,0,n+1){
        if(vis[i] == 1){
            ansmin++;
        }
    }    
    cout<<ansmin<<sp<<ansmax<<endl;
}