[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",n - ans);return 0;}

转载于:https://www.cnblogs.com/evenbao/p/9794718.html

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