ACM/ICPC World Finals 2013 F Low Power
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题目大意
有n个机器,每个机器有两块芯片,每块芯片上有k块电池,定义芯片的能量为其上k块电池的最小值,求每个机器上两块芯片的能量差的最大值的最小值
解答
很容易发现,两块芯片上电池的最小值是两个大小关系相邻的值为最优策略,然后二分答案求最小值即可。
代码:
#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <algorithm>using namespace std;int n, m, k;int e[1000500];void readin(){ scanf("%d %d", &n, &k); m = 2*n*k; for (int i = 0; i < m; i++) scanf("%d", &e[i]); sort(e, e+m);}bool check(int delta){ int cnt = 0; for (int i = m-1; i > 0; i--) if (e[i] - e[i-1] <= delta) if (m-i+1 >= 2*(cnt+1)*k) { cnt++; i--; if (cnt == n) return true; } return false;}void work(){ int l = 0, r = 1000000001; int mid; while (l <= r) { mid = (l+r) >> 1; if (check(mid)) r = mid-1; else l = mid + 1; } printf("%d", r+1);}int main(){ readin(); work(); return 0;}
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