防线问题

After the last war devastated your country, you - as the king of the land of Ardenia - decided it was high time to improve the defense of your capital city. A part of your fortification is a line of mage towers, starting near the city and continuing to the northern woods. Your advisors determined that the quality of the defense depended only on one factor: the length of a longest contiguous tower sequence of increasing heights. (They gave you a lengthy explanation, but the only thing you understood was that it had something to do with firing energy bolts at enemy forces).

After some hard negotiations, it appeared that building new towers is out of question. Mages of Ardenia have agreed to demolish some of their towers, though. You may demolish arbitrary number of towers, but the mages enforced one condition: these towers have to be consecutive.

For example, if the heights of towers were, respectively, 5, 3, 4, 9, 2, 8, 6, 7, 1, then by demolishing towers of heights 9, 2, and 8, the longest increasing sequence of consecutive towers is 3, 4, 6, 7.

Input 

Z25, denoting the number of test cases. Then Z test cases follow, each conforming to the format described below.

The input instance consists of two lines. The first one contains one positive integer n2 ^. 10⁵ denoting the number of towers. The second line containsn positive integers not larger than 10⁹ separated by single spaces being the heights of the towers.

Output 

You should output one line containing the length of a longest increasing sequence of consecutive towers, achievable by demolishing some consecutive towers or no tower at all.

Sample Input 

2 
9 
5 3 4 9 2 8 6 7 1 
7 
1 2 3 10 4 5 6

Output 

4 
6

#include"iostream"
#include"algorithm"
#include"cstring"
#include"cstdio"
using namespace std;

const int maxn=200000+10;
const int Max=1000000000;

int a[maxn],righ[maxn],lef[maxn],d[maxn],g[maxn];
int main()
{
int T;
cin>>T;
while(T--)
{
int n;
cin>>n;

for(int i=0;i<n;i++) cin>>a[i];

memset(righ,0,sizeof(righ));
memset(lef,0,sizeof(lef));

lef[0]=1;righ[n-1]=1;
for(int i=n-2;i>=0;i--)
{
if(a[i]>=a[i+1]) righ[i]=1;
else righ[i]=righ[i+1]+1;
}
for(int j=1;j<n;j++)
{
if(a[j]>a[j-1]) lef[j]=lef[j-1]+1;
else lef[j]=1;
}

//为节省数一数所耗时间而做好的预处理表格，时间复杂度从O(n^3)可降到O(n^2)，因为每次得到长度的时间从O(n)降到O(1)
/*for(int j=0;j<n;j++) cout<<righ[j]<<" ";
cout<<endl;
for(int j=0;j<n;j++) cout<<lef[j]<<" ";
cout<<endl;*/

int ans=-1;

//方法一：超时，耗时O(n^2)
/*for(int i=0;i<n;i++)
for(int j=i;j<n;j++)
if(min(lef[j]+righ[i]-1,j-i)>ans) ans=min(lef[j]+righ[i]-1,j-i);*/

//方法二：只遍历一次i，j完全靠二分查找,耗时O(nlogn)
for(int i=1;i<=n;i++) g[i]=Max;

for(int i=0;i<n;i++)
{
int k=lower_bound(g+1,g+n+1,a[i])-g; //找出的k肯定大于i，且k-i大于k+j-1.
d[i]=k+righ[i]-1; //k即为从i开始的最大递增序列长度
if(a[i]<g[lef[i]]) g[lef[i]]=a[i]; //维护左值有序表
ans=max(ans,d[i]);
}

cout<<ans<<endl;
}
return 0;
}