hdu4604 deque

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=4604

思路：就是模拟一下，求每一个开始的非上升和非下降序列。然后求重复的数，由于求出来可能不会是我们想要的序列如22221122233

这样的话求出来的会是2222222 222222233，而我们想要的到的是112222，所以不能用普通的，而是用N*logN（当然也是因为数列太长），而我们得到的是1122222多处一个2这个2是11之前留下的，但是因为2比1大才留下，那么2一定会留在最长上升里面，所以2是会被算作重复的减掉。

代码：

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <algorithm>
 #include <cstring>
 #include <map>
 #define MAXN 100010
 #define INF 1000000007
 #define FI first
 #define SE second
 using namespace std;
 int a[MAXN],f1[MAXN],f2[MAXN],d[MAXN];
 int bs(int key,int len)
 {
     int l,r;
     l = ,r = len;
     while(l <= r)
     {
         int mid = (l+r)>>;
         if(key >= f1[mid-] && key < f1[mid]) {
             //puts("yes");
             return mid;
         }
         else if(key < f1[mid]) r = mid-;
         else
         l = mid+;
     }
 }
 int bx(int key,int len)
 {
     int l,r;
     l = ,r = len;
     while(l <= r)
     {
         int mid = (l+r)>>;
         if(key <= f2[mid-] && key > f2[mid]) return mid;
         else if(key > f2[mid]) r = mid-;
         else
         l = mid+;
     }
 }
 int searchlap(int b[],int val,int len,int s,int t)
 {
     int l,r,mid;
     l = s,r = t;
     mid = (l+r)/;
     if(b[l] == val && b[r] == val) return r-l+;
     else if(l >= r) return ;
     else if(b[mid] == val) return +searchlap(b,val,len,l,mid-)+searchlap(b,val,len,mid+,r);
     else if(b[mid] > val) return searchlap(b,val,len,l,mid-);
     else if(b[mid] < val) return searchlap(b,val,len,mid+,r);
 }
 int searchlap1(int b[],int val,int len,int s,int t)
 {
     int l,r,mid;
     l = s,r = t;
     mid = (l+r)/;
     if(b[l] == val && b[r] == val) return r-l+;
     else if(l >= r) return ;
     else if(b[mid] == val) return +searchlap1(b,val,len,l,mid-)+searchlap1(b,val,len,mid+,r);
     else if(val > b[mid]) return searchlap1(b,val,len,l,mid-);
     else if (b[mid] > val)return searchlap1(b,val,len,mid+,r);
 }
 int main()
 {
     int ncase,n,ans,i,j;
     //freopen("text.txt","r",stdin);
     scanf("%d",&ncase);
     while(ncase--)
     {
         ans=;
         scanf("%d",&n);
         int len1,len2;
         for( i=;i<n;++i)    scanf("%d",&a[i]);
         f1[] = -;
         f2[] = ;
         len1 = len2 = ;
         int nowl1,nowl2;
         int cnt = ;
         int lap1,lap2;
         for(i = n-;i >= ;i--)
         {
             cnt++;
             int k;
             if(a[i] < f1[]) j = ;
             else if(a[i] >= f1[len1]) len1++,j = len1;
             else
             {
                 j = bs(a[i],len1);
             }
             f1[j] = a[i];
             lap1 = searchlap(f1,a[i],len1,,len1);
             nowl1 = j;
             if(a[i] > f2[]) j = ;
             else if(a[i] <= f2[len2]) len2++,j=len2;
             else
             {
                 j  = bx(a[i],len2);
             }
             f2[j] = a[i];
             lap2 = searchlap1(f2,a[i],len2,,len2);
             nowl2 = j;
             ans = max(ans,nowl1+nowl2-min(lap1,lap2));
         }
         cout<<ans<<endl;
     }
     return ;
 }