[12.19] Diary

12.19

DP

1. P1091: find the highest chorus formation.

Thinking : dp1 [i] i is represented by the last one, 1-i in the LIS, dp2 [i] i is represented by a first individual, in the LDS. run front to back dp1, dp2 running from the back. Note, not front to back of the LDS.

#include<bits/stdc++.h>
using namespace std;
const int M=140;
int a[M],n,ans[M];
struct LIS{
    int dp[M],d[M];
    void run(){
        int len=0;
        for(int i=1;i<=n;++i){
            int p=lower_bound(d+1,d+len+1,a[i])-d;//找到第一个大于等于a[i]的下标
            dp[i]=p,d[p]=a[i];
            if (p>len)
                len=p;
        }
    }
}L1;
struct LDS{
    int dp[M],d[M];
    void run(){
        int len=0;
        for(int i=n;i>=1;--i){
            int p=lower_bound(d+1,d+len+1,a[i])-d;//找到第一个大于等于a[i]的下标
            dp[i]=p,d[p]=a[i];
            if (p>len)
                len=p;
        }
    }
}L2;
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;++i)
        scanf("%d",&a[i]);
    L1.run(),L2.run();
    int ans=0;
    for(int i=1;i<=n;++i)
        ans=max(ans,L1.dp[i]+L2.dp[i]-1);
    printf("%d\n",n-ans);
    return 0;
}