P1091 [NOIP2004 提高组] 合唱队形（求升序序列的N^2复杂度的模板）

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int M = 1e5 + 1;
int a[M]; 
int f[3][M];
int main()
{
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j < i; j++) {
            if(a[j]<a[i])
            f[0][i] = max(f[0][i], f[0][j] + 1);
        }
    }
    for (int i = n; i >= 1; i--) {
        for (int j = n + 1; j > i; j--) {
            if(a[j]<a[i])
            f[1][i] = max(f[1][i], f[1][j] + 1);
        }
    }
    int mx = 0;
    for (int i = 1; i <= n; i++) {
        mx = max(f[0][i] + f[1][i] -1,mx);
    }
    cout << n-mx << endl;
}

二分写法
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int M = 1e5 + 1;
int a[M]; 
int f1[M];
int f2[M];
int l[M];
int r[M];
int n;
inline void solve1()
{
    int len=1;
    l[1]=1;
    f1[1]=a[1];
    for(int i=2;i<=n;i++){
        if(a[i]>f1[len]) f1[++len]=a[i];
        else{
            int p=lower_bound(f1+1,f1+len+1,a[i])-f1;
            f1[p]=a[i]; 
        }
        l[i]=len;
    }
}
inline void solve2()
{
    int len=1;
    r[1]=1;
    f2[1]=a[n];
    for(int i=n-1;i>=1;i--){
        if(a[i]>f2[len]) f2[++len]=a[i];
        else{
            int p=lower_bound(f2+1,f2+len+1,a[i])-f2;
            f2[p]=a[i]; 
        }
        r[i]=len;
    }
}
int main()
{
    
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    solve1();
    solve2();
    int mx=1e3;
    for(int i=1;i<=n;i++){
        mx=min(mx,n-l[i]-r[i+1]);
    }
    cout<<mx<<endl;
    
}

标签：NOIP2004,f1,f2,int,len,--,P1091,long,升序
来源： https://www.cnblogs.com/BlogBaudelaire/p/14634966.html