Codeforces 1111C Creative Snap分治+贪心

发布时间：2020-12-16 09:17:57 所属栏目：百科来源：网络整理

导读：Creative Snap C. Creative Snap time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Thanos wants to destroy the avengers base,but he needs to destroy the avengers along with their bas

Creative Snap

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?

if the current length is at least?
burn the current base. If it contains no avenger in it,it takes?

Output the minimum power needed by Thanos to destroy the avengers‘ base.

Input

The first line contains four integers?

The second line contains?

Output

Output one integer — the minimum power needed to destroy the avengers base.

~~这题刚一看觉得很水~~
，仔细一看数据范围……

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));
}

（编辑：李大同）

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