hzau 1205 Sequence Number（二分）

Posted 2020-09-15 gongpixin

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；

 

 1 #pragma comment(linker, "/STACK:1024000000,1024000000")
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cmath>
 5 #include<string>
 6 #include<queue>
 7 #include<algorithm>
 8 #include<stack>
 9 #include<cstring>
10 #include<vector>
11 #include<list>
12 #include<set>
13 #include<map>
14 using namespace std;
15 #define ll long long
16 #define pi (4*atan(1.0))
17 #define eps 1e-4
18 #define bug(x)  cout<<"bug"<<x<<endl;
19 const int N=1e5+10,M=1e6+10,inf=2147483647;
20 const ll INF=1e18+10,mod=2147493647;
21  
22 int a[N],nex[N];
23 int main()
24 {
25     int n;
26     while(~scanf("%d",&n))
27     {
28         memset(nex,0,sizeof(nex));
29         for(int i=1;i<=n;i++)
30             scanf("%d",&a[i]);
31         for(int j=n;j>=1;j--)
32             nex[j]=max(a[j],nex[j+1]);
33         int ans=0;
34         for(int i=1;i<=n;i++)
35         {
36             int s=i,e=n,pos=-1;
37             while(s<=e)
38             {
39                 int mid=(s+e)>>1;
40                 if(nex[mid]>=a[i])
41                     pos=mid,s=mid+1;
42                 else e=mid-1;
43             }
44             ans=max(ans,pos-i);
45         }
46         printf("%d\n",ans);
47     }
48     return 0;
49 }

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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。

 1 #include <cstdio>
 2 #include <algorithm>
 3 using namespace std;
 4 struct jj
 5 {
 6     int pos,x;
 7 }a[50000];
 8 int main()
 9 {
10     int n,i,i1,x,top,maxnum;
11     while(scanf("%d",&n)!=EOF)
12     {
13         top=0;
14         maxnum=0;
15         for(i=0;n>i;i++)
16         {
17             scanf("%d",&x);
18             if(top==0||a[top-1].x>x)
19             {
20                 a[top].x=x;
21                 a[top].pos=i;
22                 top++;
23             }
24             else
25             {
26                 for(i1=top-1;i1>=0&&a[i1].x<=x;i1--)
27                 {
28                     maxnum=max(maxnum,i-a[i1].pos);
29                 }
30             }
31         }
32         printf("%d\n",maxnum);
33     }
34     return 0;
35 }

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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

 1 #include <cstdio>
 2 #include <vector>
 3 #include <cstring>
 4 #include <string>
 5 #include <cstdlib>
 6 #include <iostream>
 7 #include <map>
 8 #include <cmath>
 9 #include <algorithm>
10 using namespace std;
11 typedef long long LL;
12 typedef pair<int,int>pii;
13 const int N = 1e5+5;
14 const double eps = 1e-8;
15 int T,n,w[N],sum[N<<2],p[N<<2],cnt,m,ret[N];
16 int k,a[N],mi[N];
17 int main() {
18     while(~scanf("%d",&n)){
19  
20         int ans=0;
21         mi[0]=1e9+10;
22         for(int i=1;i<=n;i++){
23             scanf("%d",&a[i]);
24             mi[i]=1e9+10;
25         }
26         for(int i=1;i<=n;i++){
27             mi[i]=min(mi[i-1],a[i]);
28         }
29         for(int l=n,r=n;l>=1;l--){
30             if(mi[l]>a[r]){
31                 while(a[r]<mi[l]){
32                     r--;
33                 }
34                 ans=max(ans,r-l);
35             }
36             else {
37                 ans=max(ans,r-l);
38             }
39         }
40         printf("%d\n",ans);
41     }
42    return 0;
43 }

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