最长上升子序列1006

Removed Interval

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 65536/65536K (Java/Other)

Total Submission(s) : 4   Accepted Submission(s) : 2

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

Problem Description

Given a sequence of numbers A=a1,a2,…,aN, a subsequence b1,b2,…,bk of A is referred as increasing if b1<b2<…<bk. LY has just learned how to find the longest increasing subsequence (LIS).
Now that he has to select L consecutive numbers and remove them from A for some mysterious reasons. He can choose arbitrary starting position of the selected interval so that the length of the LIS of the remaining numbers is maximized. Can you help him with this problem?

Input

The first line of input contains a number T indicating the number of test cases (T≤100).
For each test case, the first line consists of two numbers N and L as described above (1≤N≤100000,0≤L≤N). The second line consists of N integers indicating the sequence. The absolute value of the numbers is no greater than 109.
The sum of N over all test cases will not exceed 500000.

Output

For each test case, output a single line consisting of “Case #X: Y”. X is the test case number starting from 1. Y is the maximum length of LIS after removing the interval.

Sample Input

25 21 2 3 4 55 35 4 3 2 1

Sample Output

Case #1: 3Case #2: 1

思路：这题如果直接枚举删除的话再找出最长的上升子序列这样时间会超出，然后可以找出要删除的左右两边最大上升子序列之和，以代替枚举要删除的地方

之后的最长上升子序列。

代码：

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#define INF 0x3f3f3fusing namespace std;int a[100005];int b[100005];int c[100005];int dp[100005];int main(){    int T;    cin>>T;    int t=1;    while(T--)    {        int m,n;        cin>>m>>n;        int i,x;        for(i=0;i<m;i++)        {            scanf("%d",&a[i]);            b[i]=-a[i];        }         memset(c, 0x7f, sizeof(c));        for(i=m-1;i>=0;i--)        {            x=lower_bound(c,c+m,b[i])-c;            c[x]=b[i];            dp[i]=x+1;        }         memset(c, 0x7f, sizeof(c));        int zz=0,len=0;        for(i=n;i<m;i++)        {            x=lower_bound(c,c+m,a[i])-c;            zz=max(zz,x+1+dp[i]-1);            x=lower_bound(c,c+m,a[i-n])-c;            c[x]=a[i-n];            len=max(len,x+1);        }        printf("Case #%d: ", t++);        if(zz>len)len=zz;        cout<<len<<endl;    }    return 0;}