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 2
. 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:
, divide the base into 2
amount of power, otherwise it takes his B?na?l amount of power, where na is the number of avengers and l
The first line contains four integers n
, k, A and B (1≤n≤30, 1≤k≤105, 1≤A,B≤104), where 2n is the length of the base, k is the number of avengers and A and B
are the constants explained in the question.
The second line contains k
integers a1,a2,a3,…,ak (1≤ai≤2n), where ai
represents the position of avenger in the base.
Output one integer — the minimum power needed to destroy the avengers base.
2 2 1 2 1 3
6
3 2 1 2 1 7
8
Consider the first example.
One option for Thanos is to burn the whole base 1?4
with power 2?2?4=16
.
Otherwise he can divide the base into two parts 1?2
and 3?4
.
For base 1?2
, he can either burn it with power 2?1?2=4 or divide it into 2 parts 1?1 and 2?2
.
For base 1?1
, he can burn it with power 2?1?1=2. For 2?2, he can destroy it with power 1, as there are no avengers. So, the total power for destroying 1?2 is 2+1=3, which is less than 4
.
Similarly, he needs 3
power to destroy 3?4. The total minimum power needed is 6.
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<time.h>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 1000005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline int rd() {
int x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == ‘-‘) f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
int n, k;
int A, B;
int a[maxn];
ll dfs(ll l, ll r) {
ll L = lower_bound(a + 1, a + 1 + k, l) - a;
ll R = upper_bound(a + 1, a + 1 + k, r) - a; R--;
ll tot = R - L + 1;
ll cost = 0;
if (tot == 0)cost = A;
else cost = B * (r - l + 1)*tot;
if (l == r || tot == 0)return cost;
ll mid = (l + r) >> 1;
return min(cost, dfs(l, mid) + dfs(mid+1, r));
}
int main()
{
// ios::sync_with_stdio(0);
rdint(n); rdint(k); rdint(A); rdint(B);
for (int i = 1; i <= k; i++)rdint(a[i]);
sort(a + 1, a + 1 + k);
ll x = (ll)(1 << n);
cout << (ll)dfs(1, x) << endl;
return 0;
}
CodeCraft-19 and Codeforces Round #537 (Div. 2) C. Creative Snap 分治
原文：https://www.cnblogs.com/zxyqzy/p/10359275.html