ZOJ4107 Singing Everywhere

给出n个数 最多可以消除一个数 问这个n或n-1个数中满足a[i] > a[i - 1] && a[i] > a[i + 1]的数最少

直接模拟就行了

AC代码：

 1 #include<bits/stdc++.h>
 2 #define pi acos(-1)
 3 #define INF 0x3f3f3f3f
 4 typedef long long ll;
 5 typedef unsigned long long ull;
 6 using namespace std;
 7 
 8 namespace io {
 9     const int SIZE = 1e7 + 10;
10     char inbuff[SIZE];
11     char *l, *r;
12     inline void init() {
13         l = inbuff;
14         r = inbuff + fread(inbuff, 1, SIZE, stdin);
15     }
16     inline char gc() {
17         if (l == r) init();
18         return (l != r) ? *(l++) : EOF;
19     }
20     void read(int &x) {
21         x = 0; char ch = gc();
22         while(!isdigit(ch)) ch = gc();
23         while(isdigit(ch)) x = x * 10 + ch - '0', ch = gc();
24     }
25 } using io::read;
26 
27 bool cmp(const int &a, const int &b){
28     return a > b;
29 }
30 
31 int main(){
32     ios::sync_with_stdio(false);
33     int t;
34     cin>>t;
35     ll a[100005];
36     int n;
37     while (t--){
38         cin>>n;
39         memset(a, 0, sizeof(a));
40         for (int i = 1; i <= n; i++) cin>>a[i];
41         a[0] = 9223372036854775807, a[n + 1] = 9223372036854775807;
42         int cnt = 0;
43         for (int i = 2; i < n; i++)
44             if (a[i] > a[i - 1] && a[i] >a[i + 1])
45                 cnt++;
46         int ans = 0;
47         int ans1, ans2;
48         for (int i = 2; i < n; i++){
49             ans1 = 0, ans2 = 0;
50             if (a[i - 1] > a[i + 1] && a[i - 1] > a[i - 2]) ans1++;
51             if (a[i + 1] > a[i - 1] && a[i + 1] > a[i + 2]) ans1++;
52             if (a[i - 1] > a[i - 2] && a[i - 1] > a[i]) ans2++;
53             if (a[i + 1] > a[i + 2] && a[i + 1] > a[i]) ans2++;
54             if (a[i] > a[i - 1] && a[i] > a[i + 1]) ans2++;
55             ans = max(ans, ans2 - ans1);
56         }
57         cout<<cnt - ans<<endl;
58     }
59     return 0;
60 }