https://vjudge.net/contest/161167#problem/F——dp

Think:
1求非严格单调递增子序列和非严格单调递减子序列
2反思：
1>lower_bound()—-Wrong Answer
upper_bound()—-Accepted

建议参考博客

以下为Wrong Answer代码——lower_bound()可行数据覆盖？

#include 
#include 

using namespace std;

int n, a[100004], b[100004];

int dp1();
int dp2();

int main(){
    int T, i;
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        for(i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        int x1 = dp1();
        int x2 = dp2();
        if(x1 == n-1 || x1 == n || x2 == n-1 || x2 == n)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}
int dp1(){
    int i, len, pos;
    b[1] = a[1];
    len = 1;
    for(i = 2; i <= n; i++){
        if(a[i] >= b[len]){
            len += 1;
            b[len] = a[i];
        }
        else {
            pos = lower_bound(b+1, b+1+len, a[i]) - b;
            b[pos] = a[i];
        }
    }
    return len;
}
int dp2(){
    int i, len, pos;
    b[1] = a[n];
    len = 1;
    for(i = n-1; i >= 1; i--){
        if(a[i] >= b[len]){
            len += 1;
            b[len] = a[i];
        }
        else {
            pos = lower_bound(b+1, b+1+len, a[i]) - b;
            b[pos] = a[i];
        }
    }
    return len;
}

以下为Accepted代码

#include 
#include 

using namespace std;

int n, a[100004], b[100004];

int dp1();
int dp2();
int s(int num, int l, int h);

int main(){
    int T, i;
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        for(i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        int x1 = dp1();
        int x2 = dp2();
        if(x1 == n-1 || x1 == n || x2 == n-1 || x2 == n)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}
int dp1(){
    int i, len, pos;
    b[1] = a[1];
    len = 1;
    for(i = 2; i <= n; i++){
        if(a[i] >= b[len]){
            len += 1;
            b[len] = a[i];
        }
        else {
            pos = upper_bound(b+1, b+len+1, a[i]) - b;
            //pos = lower_bound(b+1, b+len+1, a[i]) - b;
            //pos = s(a[i], 1, len);
            b[pos] = a[i];
        }
    }
    return len;
}
int dp2(){
    int i, len, pos;
    b[1] = a[n];
    len = 1;
    for(i = n-1; i >= 1; i--){
        if(a[i] >= b[len]){
            len += 1;
            b[len] = a[i];
        }
        else {
            pos = upper_bound(b+1, b+len+1, a[i]) - b;
            //pos = lower_bound(b+1, b+len+1, a[i]) - b;
            //pos = s(a[i], 1, len);
            b[pos] = a[i];
        }
    }
    return len;
}
int s(int num,int l,int h)  
{  
    int mid;  
    while(l<=h)  
    {  
        mid=(l+h)/2;  
        if(num>=b[mid])  
            l=mid+1;  
        else h=mid-1;  
    }  
    return l;  
}