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  • 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 }
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  • 原文地址:https://www.cnblogs.com/Asurudo/p/11319772.html
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