D. Psychos in a Line Codeforces Round #189 (Div. 2)

D. Psychos in a Linetime limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n psychos standing in a line. Each psycho is assigned a unique integer from 1 to n.
At each step every psycho who has an id greater than the psycho to his right (if exists) kills his right neighbor in the line. Note that a psycho might kill and get killed at the same step.

You're given the initial arrangement of the psychos in the line. Calculate how many steps are needed to the moment of time such, that nobody kills his neighbor after that moment. Look notes to understand the statement more precise.

Input

The first line of input contains integer n denoting the number of psychos, (1 ≤ n ≤ 10^{5).
In the second line there will be a list of n space separated distinct integers each in range 1 to n,
inclusive — ids of the psychos in the line from left to right.}

Output

Print the number of steps, so that the line remains the same afterward.

Sample test(s)input

1010 9 7 8 6 5 3 4 2 1

output

2

input

61 2 3 4 5 6

output

0

Note

In the first sample line of the psychos transforms as follows: [10 9 7 8 6 5 3 4 2 1]  →  [10 8 4]  →  [10].
So, there are two steps.

思路：考慮第i個人是被誰殺的。他可以被他前一個人殺，如果殺不了，則被殺掉前一個人的人所殺，類推，找到能殺掉第i個人的人，若都殺不死他，則記錄不能殺死他。

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>using namespace std;int n,ans,flag;int a[100005],b[100005],xb[100005]; // a[]-存輸入的資料//  b[]-存需要幾次操作將他殺掉 xb[]-存殺掉第i個人的人的下標void solve(){    int i,j,ma;    ans=-1;    b[1]=xb[1]=0;     // 預設b[1]殺不掉    for(i=2; i<=n; i++)  //從第二個開始判斷    {        ma=0;        flag=0;     // 存a[i]是否能被殺掉        j=i-1;        while(a[j]<a[i]&&j)        {            if(!b[j])  // 如果a[j]不能被前面人殺掉 則a[i]也不能            {                flag=1;                break;            }            if(b[j]>ma) ma=b[j];            j=xb[j];   // 跳轉到殺死a[j]的人        }        if(!flag)   b[i]=ma+1;  //如果被殺死 則需ma+1步        else   b[i]=0;          // 沒被殺死則需要0步        xb[i]=j;       // 記錄是誰殺死他的//     printf("i:%d b[i]:%d xb[i]:%d\n",i,b[i],xb[i]);    }    for(i=1; i<=n; i++)    {        if(ans<b[i]) ans=b[i];    }}int main(){    int i,j;    while(~scanf("%d",&n))    {        for(i=1; i<=n; i++)        {            scanf("%d",&a[i]);        }        solve();        printf("%d\n",ans);    }    return 0;}