hdu 5489 Removed Interval 动态规划，最长上升子序列

Removed Interval

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 849    Accepted Submission(s): 310

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

 

Source

2015 ACM/ICPC Asia Regional Hefei Online

 

假设最长子序列在被删除位置的起点的左边，即满足 i < n - L

那么答案就是【0，i】中i可以排到的最长的位置

假设删除区间在最长子序列中间部分

枚举每一个点x，删除位置为[x-l-1,x-1]。

那么计算出[0,x-l-1]的最长上升序列的情况，然后看第x个数能加入这些序列最长的位置y。那么答案就是

y + x能够出现在[x,n]区间中以x开头的最长子序列的长度（相当于逆序求x所在最长下降子序列的的长度）。

证明如下：

     对于一个x，删除[x-l-1,x-1]的位置得到的结果最优。

    显然对于x来说得到的结果是[0,x-l-1]+[x,n]包含x的最长子序列长度。

    那么可能的情况是 存在y，y的位置在[x-l-1,x-1]区间，且x,y同时存在这个最长的子序列结果中。

    那么显然在[y,n]区间x会被计算到关于y必选的子序列的情况中。并且关于y的最长的子序列的长度已经在之前计算得到。

#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>using namespace std;#define maxn 100007int dp[maxn];int rs[maxn],ls[maxn];int num[maxn];int find(int n,int val){    int low = 0,high = n-1;    while(low <= high){        int mid = (low+high)/2;        if(dp[mid] >= val) high = mid-1;        else low = mid+1;    }    return low;}int main(){    int t,n,l,tt=1;    scanf("%d",&t);    while(t--){        scanf("%d%d",&n,&l);        for(int i = 0;i < n; i++)            scanf("%d",&num[i]);        int cnt = 0;        for(int i = n-1;i >= 0; i--){            int p = find(cnt,-num[i]);            dp[p] = -num[i];            rs[i] = p;            if(p == cnt) cnt++;        }        cnt = 0;        int ans = 0;        memset(ls,0,sizeof(ls));        for(int i = 0;i < n; i++){            if(i+l < n){                int p = find(cnt,num[i+l]);                ls[i+l] = p;            }            int p = find(cnt,num[i]);            dp[p] = num[i];            if(p == cnt) cnt++;            if(i < n - l) ans = max(ans,p+1);        }        for(int i = l;i < n; i++){            ans = max(ans,rs[i]+ls[i]+1);        }        printf("Case #%d: %d\n",tt++,ans);    }    return 0;}