P1091-合唱队形

 1 #include <bits/stdc++.h>
 2 #define _for(i,a,b) for(int i = (a);i < b;i ++)
 3 typedef long long ll;
 4 using namespace std;
 5 inline ll read()
 6 {
 7     ll ans = 0;
 8     char ch = getchar(), last = ' ';
 9     while(!isdigit(ch)) last = ch, ch = getchar();
10     while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
11     if(last == '-') ans = -ans;
12     return ans;
13 }
14 inline void write(ll x)
15 {
16     if(x < 0) x = -x, putchar('-');
17     if(x >= 10) write(x / 10);
18     putchar(x % 10 + '0');
19 }
20 int N;
21 int a[103];
22 
23 int sv1(int k)
24 {
25     int dp1[103];
26     int dp1ed = 1;
27     dp1[dp1ed] = a[0];
28     _for(i,1,k+1)
29         if(a[i]>dp1[dp1ed])
30             dp1[++dp1ed] = a[i];
31         else
32         {
33             int t = lower_bound(dp1+1,dp1+dp1ed+1,a[i])-dp1;
34             dp1[t] = a[i];
35         }
36     return k+1-dp1ed;
37 }
38 int sv2(int k)
39 {
40     int dp2[103];
41     int dp2ed = 1;
42     dp2[dp2ed] = a[k];
43     _for(i,k,N)
44         if(a[i]<dp2[dp2ed])
45             dp2[++dp2ed] = a[i];
46         else
47         {
48             int t = lower_bound(dp2+1,dp2+dp2ed+1,a[i],greater<int>())-dp2;
49             dp2[t] = a[i];
50         }
51     return N-k-dp2ed;
52 }
53 int main()
54 {
55     N = read();
56     _for(i,0,N)
57         a[i] = read();
58     
59     int rnt = 200;
60     _for(i,0,N)
61     {
62         int rt1 = sv1(i);
63         int rt2 = sv2(i);
64         rnt = min(rnt,rt1+rt2);
65     }    
66     write(rnt);
67     return 0;
68 }

标签：ch,dp2,dp1,合唱队,int,P1091,rnt,dp1ed
来源： https://www.cnblogs.com/Asurudo/p/11319772.html