2016 EC-Final Problem D. Ice Cream Tower(二分+贪心)
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题目地址:点击打开链接
题意:有N个冰淇淋球,做一个冰淇淋需要K个球,这K个球还必须满足上下相邻的下面比上面大至少两倍。先给出N个球的质量,问最多能做出多少个冰淇淋?
思路:二分可以做的冰淇淋数,贪心验证是否可以堆成,将冰淇淋按从小到大排序,一层一层贪心地堆就可以了,因为
这个冰淇淋如果当前不能用上,那么后续更不可能用上。
代码:
#include<bits/stdc++.h>using namespace std;typedef long long ll;const int maxn = 1e6+5;int n, k;ll a[maxn], rec[maxn][2];bool judge(int x){ int pre = 0, now = 1, p = 1; for(int i = 1; i <= x; i++) rec[i][pre] = a[p++]; for(int i = 1; i < k; i++) { for(int j = 1; j <= x; j++) { while(p <= n && a[p] < rec[j][pre]*2) p++; if(p > n) return 0; rec[j][now] = a[p++]; } swap(pre, now); } return 1;}int main(void){ int t, ca = 1; cin >> t; while(t--) { scanf("%d%d", &n, &k); for(int i = 1; i <= n; i++) scanf("%lld", &a[i]); sort(a+1, a+1+n); int l = 0, r = n/k, ans; while(l <= r) { int mid = (l+r)/2; if(judge(mid)) ans = mid, l = mid+1; else r = mid-1; } printf("Case #%d: %d\n", ca++, ans); } return 0;}
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