标签：注意 math exp plain ssis mount hal either and
Thanos wants to destroy the avengers base, but he needs to destroy the avengers along with their base.
Let we represent their base with an array, where each position can be occupied by many avengers, but one avenger can occupy only one position. Length of their base is a perfect power of 22. Thanos wants to destroy the base using minimum power. He starts with the whole base and in one step he can do either of following:
The first line contains four integers nn, kk, AA and BB (1≤n≤301≤n≤30, 1≤k≤1051≤k≤105, 1≤A,B≤1041≤A,B≤104), where 2n2n is the length of the base, kkis the number of avengers and AA and BB are the constants explained in the question.
The second line contains kk integers a1,a2,a3,…,aka1,a2,a3,…,ak (1≤ai≤2n1≤ai≤2n), where aiai represents the position of avenger in the base.
Output one integer — the minimum power needed to destroy the avengers base.
这题刚一看觉得很水
其实也不难，主要是数据范围过大。要用一种类似离散化的做法。
对数组a排序。对于一段区间[l,r]的英雄数量等于a中第一个比r大的数的下标减1减a中第一个大于等于l的数的下标加1，这样就可以做了(k很小)。
不过要注意剪枝：若区间[l,r]的英雄数量等于0，就直接返回A。
上代码：
#include <bits/stdc++.h> using namespace std; long long n, k, a, b; long long hero[1000001]; long long solve(long long l, long long r) { long long num = upper_bound(hero + 1, hero + 1 + k, r) - lower_bound(hero + 1, hero + 1 + k, l); if (num == 0) return a; long long ans = (r - l + 1) * b * num; if (l >= r) return ans; ans = min(ans, solve(l, (l + r) / 2) + solve((l + r) / 2 + 1, r)); return ans; } int main() { cin >> n >> k >> a >> b; for (long long i = 1; i <= k; i++) cin >> hero[i]; sort(hero + 1, hero + 1 + k); cout << solve(1, (1 << n)); }
Codeforces 1111C Creative Snap分治+贪心
标签：注意 math exp plain ssis mount hal either and
原文地址：https://www.cnblogs.com/zcr-blog/p/11427268.html
