给你一个数组，最多可以修改一个数，问最少的山峰和山谷数量之和。

小菜鸡做了一年。
一上来想猜个结论，让每个数等于其相邻的两个数，看了题解之后也证明的正确性，当时直接写就没后面这么多事了。
但是命运让我分情况讨论(逃。
画了一张纸的图，让后面向样例编程。终于是把分类讨论的方法给过掉了。。。太麻烦就不讲啦，直接贴一个题解的思路的代码吧。

//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n;
int a[N],c[N];bool get(int a[],int x)
{if(x<=1||x>=n) return 0;if(a[x]>a[x-1]&&a[x]>a[x+1]||a[x]<a[x-1]&&a[x]<a[x+1]) return 1;return 0;
}int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);int _; scanf("%d",&_);while(_--){scanf("%d",&n);for(int i=1;i<=n;i++) scanf("%d",&a[i]),c[i]=a[i];int ans=0,sum=0;if(n<=3) { puts("0"); continue; }for(int i=2;i<=n-1;i++) if(get(a,i)) ans++,sum++;for(int i=1;i<=n;i++){int t=a[i];a[i]=a[i-1];ans=min(ans,sum-get(c,i)-get(c,i-1)-get(c,i+1)+get(a,i)+get(a,i-1)+get(a,i+1));a[i]=a[i+1];ans=min(ans,sum-get(c,i)-get(c,i-1)-get(c,i+1)+get(a,i)+get(a,i-1)+get(a,i+1));a[i]=t;}printf("%d\n",ans);}return 0;
}
/**/

分类讨论：

//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n;
int a[N],c[N];int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);//rd_wa();int _; scanf("%d",&_);while(_--){//int id; scanf("%d",&id);//printf("%d\n",id);scanf("%d",&n);for(int i=1;i<=n;i++) scanf("%d",&a[i]);// 1 谷  2峰int ans=0,sum;for(int i=2;i<=n-1;i++)if(a[i]>a[i-1]&&a[i]>a[i+1]) c[i]=2,ans++;else if(a[i]<a[i-1]&&a[i]<a[i+1]) c[i]=1,ans++;else c[i]=0;sum=ans;if(c[2]||c[n-1]) ans--;for(int i=2;i<=n-1;i++){if(c[i]){ans=min(ans,sum-1);if(c[i-1]&&!c[i+1]){if(c[i-1]==2&&a[i+1]>=a[i-1]||(i+1==n||a[i+1]==a[i+2])) ans=min(ans,sum-2);if(c[i-1]==1&&a[i+1]<=a[i-1]||(i+1==n||a[i+1]==a[i+2])) ans=min(ans,sum-2);}if(c[i+1]&&!c[i-1]){if(c[i+1]==2&&a[i-1]>=a[i+1]||(i-1==1||a[i-1]==a[i-2])) ans=min(ans,sum-2);if(c[i+1]==1&&a[i-1]<=a[i+1]||(i-1==1||a[i-1]==a[i-2])) ans=min(ans,sum-2);}}//else if(c[i-1]||c[i+1]) ans=min(ans,sum-1);if(c[i-1]&&c[i+1]&&c[i-1]==c[i+1]) ans=min(ans,sum-3);else if(c[i-1]&&c[i+1]) ans=min(ans,sum-1);}printf("%d\n",ans);for(int i=1;i<=n;i++) c[i]=0;}return 0;
}
/*
1
14
62 47 35 39 42 54 60 90 41 80 80 74 67 42
*/