hzau 1205 Sequence Number（二分）

G. Sequence Number

In Linear algebra, we have learned the definition of inversion number: Assuming A is a ordered set with n numbers ( n > 1 ) which are different from each other. If exist positive integers i , j, ( 1 ≤ i ＜ j ≤ n and A[i] ＞ A[j]), <a[i], a[j]=""> is regarded as one of A’s inversions. The number of inversions is regarded as inversion number. Such as, inversions of array <2,3,8,6,1> are <2,1>, <3,1>, <8,1>, <8,6>, <6,1>,and the inversion number is 5. Similarly, we define a new notion —— sequence number, If exist positive integers i, j, ( 1 ≤ i ≤ j ≤ n and A[i] <= A[j], <a[i], a[j]=""> is regarded as one of A’s sequence pair. The number of sequence pairs is regarded as sequence number. Define j – i as the length of the sequence pair. Now, we wonder that the largest length S of all sequence pairs for a given array A.

Input

There are multiply test cases. In each case, the first line is a number N(1<=N<=50000 ), indicates the size of the array, the 2th ~n+1th line are one number per line, indicates the element Ai （1<=Ai<=10^9） of the array.

Output

Output the answer S in one line for each case.

Sample Input

5 2 3 8 6 1

Sample Output

3

题意：找出最远的i<=j&&a[i]<=a[j]的长度；

思路：这是一道排序可以过的题，也可以rmq+二分 最快的写法可以用单调栈做到O(n)

　　　我是求后面的最大值后缀，二分后缀；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-4
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e5+10,M=1e6+10,inf=2147483647;
const ll INF=1e18+10,mod=2147493647;
 
int a[N],nex[N];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(nex,0,sizeof(nex));
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(int j=n;j>=1;j--)
            nex[j]=max(a[j],nex[j+1]);
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            int s=i,e=n,pos=-1;
            while(s<=e)
            {
                int mid=(s+e)>>1;
                if(nex[mid]>=a[i])
                    pos=mid,s=mid+1;
                else e=mid-1;
            }
            ans=max(ans,pos-i);
        }
        printf("%d
",ans);
    }
    return 0;
}

单调栈做法，如果当前元素小于栈顶元素入栈，否则依次和栈中元素计算相对距离且取最长的那一个。

#include <cstdio>
#include <algorithm>
using namespace std;
struct jj
{
    int pos,x;
}a[50000];
int main()
{
    int n,i,i1,x,top,maxnum;
    while(scanf("%d",&n)!=EOF)
    {
        top=0;
        maxnum=0;
        for(i=0;n>i;i++)
        {
            scanf("%d",&x);
            if(top==0||a[top-1].x>x)
            {
                a[top].x=x;
                a[top].pos=i;
                top++;
            }
            else
            {
                for(i1=top-1;i1>=0&&a[i1].x<=x;i1--)
                {
                    maxnum=max(maxnum,i-a[i1].pos);
                }
            }
        }
        printf("%d
",maxnum);
    }
    return 0;
}

双指针做法，维护左指针和右指针，使左指针的值小于等于右指针的值

#include <cstdio>
#include <vector>
#include <cstring>
#include <string>
#include <cstdlib>
#include <iostream>
#include <map>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int,int>pii;
const int N = 1e5+5;
const double eps = 1e-8;
int T,n,w[N],sum[N<<2],p[N<<2],cnt,m,ret[N];
int k,a[N],mi[N];
int main() {
    while(~scanf("%d",&n)){
 
        int ans=0;
        mi[0]=1e9+10;
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            mi[i]=1e9+10;
        }
        for(int i=1;i<=n;i++){
            mi[i]=min(mi[i-1],a[i]);
        }
        for(int l=n,r=n;l>=1;l--){
            if(mi[l]>a[r]){
                while(a[r]<mi[l]){
                    r--;
                }
                ans=max(ans,r-l);
            }
            else {
                ans=max(ans,r-l);
            }
        }
        printf("%d
",ans);
    }
   return 0;
}