Hills And Valleys CodeForces - 1467B

题意：

修改数列中的 一个 数字 使得峰(波峰、波谷)的数量最少

题解：

修改一个数，最多只能影响左右两个数，所能减少的峰的数量为1,2，3三种
分类讨论，对于当前位置i，如果我们将a[i]变成比左右两边都小的情况，比左右两边都大的情况，位于两者中间的情况，等于左边的情况，等于右边的情况，以上所有情况均取最大的到temp
然后用总数量减去temp

代码:

#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 10;
#define int long long
int t, n, a[maxn];
int judge(int i) {if (i >= 2 && i <= n - 1) {if (a[i] > a[i + 1] && a[i] > a[i - 1]) return 1;//山峰 if (a[i] < a[i + 1] && a[i] < a[i - 1]) return 1;//山谷 }return 0;
}
signed main() {cin >> t;while (t--) {cin >> n;for (int i = 1; i <= n; i++) cin >> a[i];int ans = 0, temp = 0;for (int i = 2; i <= n - 1; i++)if (judge(i)) ans++;for (int i = 1; i <= n; i++) {int now = judge(i - 1) + judge(i + 1) + judge(i);int last = 1e9;int f = a[i];a[i] = min(a[i - 1], a[i + 1]) - 1;  //比小的小last = min(last, judge(i - 1) + judge(i + 1) + judge(i));a[i] = max(a[i - 1], a[i + 1]) + 1;  //比大的大last = min(last, judge(i - 1) + judge(i + 1) + judge(i));a[i] = min(a[i - 1], a[i + 1]) + 1;  //位于中间last = min(last, judge(i - 1) + judge(i + 1) + judge(i));a[i] = a[i - 1];  //相等last = min(last, judge(i - 1) + judge(i + 1) + judge(i));a[i] = a[i + 1];  //相等last = min(last, judge(i - 1) + judge(i + 1) + judge(i));temp = max(temp, now - last);a[i] = f;}cout << ans - temp << endl;}
}