CodeForces-91B——Queue（单调队列之海象比年龄）

There are n walruses standing in a queue in an airport. They are numbered starting from the queue's tail: the 1-st walrus stands at the end of the queue and the n-th walrus stands at the beginning of the queue. The i-th walrus has the age equal to a_i.

The i-th walrus becomes displeased if there's a younger walrus standing in front of him, that is, if exists such j (i < j), that a_i > a_j. The displeasure of the i-th walrus is equal to the number of walruses between him and the furthest walrus ahead of him, which is younger than the i-th one. That is, the further that young walrus stands from him, the stronger the displeasure is.

The airport manager asked you to count for each of n walruses in the queue his displeasure.

The first line contains an integer n (2 ≤ n ≤ 10⁵) — the number of walruses in the queue. The second line contains integers a_i (1 ≤ a_i ≤ 10⁹).

Note that some walruses can have the same age but for the displeasure to emerge the walrus that is closer to the head of the queue needs to be strictly younger than the other one.

Print n numbers: if the i-th walrus is pleased with everything, print "-1" (without the quotes). Otherwise, print the i-th walrus's displeasure: the number of other walruses that stand between him and the furthest from him younger walrus.

6
10 8 5 3 50 45

2 1 0 -1 0 -1 

7
10 4 6 3 2 8 15

4 2 1 0 -1 -1 -1 

5
10 3 1 10 11

1 0 -1 -1 -1 

       从尾部开始遍历，单调存递减序列，二分查找。

#include
#include
#include
#define M 100010
using namespace std;
int a[M],tail,ans[M];
struct node
{
    int x,i;
};
node q[M];
int erfen(int x)
{
    int l=0,r=tail,pan;
    while(r>=l)
    {
        pan=(r+l)/2;
        if(x>q[pan].x)
        {
            r=pan-1;
        }
        else
        {
            l=pan+1;
        }
    }
    return l;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        tail=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i=n;i>0;i--)
        {
            if(tail==0||q[tail-1].x>=a[i])
            {
                q[tail].x=a[i];
                q[tail++].i=i;
                ans[i]=-1;
            }
            else
            {
                int j=erfen(a[i]);
                ans[i]=q[j].i-i-1;
            }
        }
        for(int i=1;i