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

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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 <cstdio>#include <algorithm>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 <cstdio>#include <algorithm>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;  }

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