[USACO 2017DEC] Greedy Gift Takers

[题目链接]

         https://www.lydsy.com/JudgeOnline/problem.php?id=5139

[算法]

        二分答案

        时间复杂度 : O(NlogN^2)

[代码]

        

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;

int n;
int a[MAXN] , b[MAXN];
 
template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
inline bool check(int x)
{
        for (int i = 1; i < x; i++) b[i] = a[i];
        sort(b + 1,b + x);
        int limit = n - x;
        for (int i = 1; i < x; i++)
        {
                if (b[i] > limit) return false;
                ++limit;        
        }        
        return true;
}

int main()
{
        
        read(n);
        for (int i = 1; i <= n; i++) read(a[i]);
        int l = 1 , r = n , ans = 0;
        while (l <= r)
        {
                int mid = (l + r) >> 1;
                if (check(mid))
                {
                        l = mid + 1;
                        ans = mid;        
                }    else r = mid - 1;
        }
        printf("%d
",n - ans);
        
        return 0;
    
}