[AcWing 244] 谜一样的牛
作者:互联网
树状数组 + 二分 复杂度 \(n \cdot log^{2}(n)\)
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#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 2e5 + 10;
int n;
int h[N];
int tr[N];
int ans[N];
int lowbit(int x)
{
return x & -x;
}
void add(int x, int c)
{
for (int i = x; i <= n; i += lowbit(i))
tr[i] += c;
}
int ask(int x)
{
int res = 0;
for (int i = x; i; i -= lowbit(i))
res += tr[i];
return res;
}
void solve()
{
cin >> n;
for (int i = 2; i <= n; i ++)
cin >> h[i];
for (int i = 1; i <= n; i ++)
add(i, 1);
for (int i = n; i; i --) {
int k = h[i] + 1;
int l = 1, r = n;
while (l < r) {
int mid = l + r >> 1;
if (ask(mid) >= k)
r = mid;
else
l = mid + 1;
}
ans[i] = l;
add(l, -1);
}
for (int i = 1; i <= n; i ++)
cout << ans[i] << endl;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
solve();
return 0;
}
- 用 \(tr[i]\) 记录数字 \(i\) 是否被用过,\(tr[i] = 1\) 表示没有被用过,\(tr[i] = 0\) 表示被用过,用树状数组维护前缀和
- 从后往前进行,第 \(i\) 头牛前面有 \(h[i]\) 头牛比它低,那么它是第 \(h[i] + 1\) 高,在树状数组的前缀和中二分,找到第一个大于等于 \(h[i] + 1\) 的位置
标签:树状,谜一样,tr,mid,244,int,数组,ans,AcWing 来源: https://www.cnblogs.com/wKingYu/p/16496264.html