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+,M=1e6+,inf=;
 const ll INF=1e18+,mod=;

 int a[N],nex[N];
 int main()
 {
     int n;
     while(~scanf("%d",&n))
     {
         memset(nex,,sizeof(nex));
         for(int i=;i<=n;i++)
             scanf("%d",&a[i]);
         for(int j=n;j>=;j--)
             nex[j]=max(a[j],nex[j+]);
         int ans=;
         for(int i=;i<=n;i++)
         {
             int s=i,e=n,pos=-;
             while(s<=e)
             {
                 int mid=(s+e)>>;
                 if(nex[mid]>=a[i])
                     pos=mid,s=mid+;
                 else e=mid-;
             }
             ans=max(ans,pos-i);
         }
         printf("%d\n",ans);
     }
     return ;
 }

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[];
 int main()
 {
     int n,i,i1,x,top,maxnum;
     while(scanf("%d",&n)!=EOF)
     {
         top=;
         maxnum=;
         for(i=;n>i;i++)
         {
             scanf("%d",&x);
             if(top==||a[top-].x>x)
             {
                 a[top].x=x;
                 a[top].pos=i;
                 top++;
             }
             else
             {
                 for(i1=top-;i1>=&&a[i1].x<=x;i1--)
                 {
                     maxnum=max(maxnum,i-a[i1].pos);
                 }
             }
         }
         printf("%d\n",maxnum);
     }
     return ;
 }

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+;
 const double eps = 1e-;
 int T,n,w[N],sum[N<<],p[N<<],cnt,m,ret[N];
 int k,a[N],mi[N];
 int main() {
     while(~scanf("%d",&n)){

         int ans=;
         mi[]=1e9+;
         for(int i=;i<=n;i++){
             scanf("%d",&a[i]);
             mi[i]=1e9+;
         }
         for(int i=;i<=n;i++){
             mi[i]=min(mi[i-],a[i]);
         }
         for(int l=n,r=n;l>=;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\n",ans);
     }
    return ;
 }