Removed Interval
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2223 Accepted Submission(s): 700
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
2 5 2 1 2 3 4 5 5 3 5 4 3 2 1
Sample Output
Case #1: 3 Case #2: 1
Source
2015 ACM/ICPC Asia Regional Hefei Online
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#include<bits/stdc++.h>
using namespace std;#define debug puts("YES");
#define rep(x,y,z) for(int (x)=(y);(x)<(z);(x)++)
#define read(x,y) scanf("%d%d",&x,&y)#define root l,r,rt
#define lrt int l,int r,int rt
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
const int maxn =5e5+5;
/*
题目大意:给定一个序列，要求其删除连续k个后的最长
LIS长度。首先看题解才知道离散化数据的概念，
也是张了见识了，离散化数据在线段树构建中很常用，
就是把不对数的绝对大小进行建树，而对数的大小序号序列进行建树。在滑动窗口滑动的时候，每次都不断的更新左断扫过的数据，更新到ST中，
更新的位置是离散化的位置，这样每次查询的时候就可以直接引用全局最大值，
因为破坏查询位置答案的数字还没出现，或者是更新不到查询区间中，究其本质
还是大小关系，查询要求比目标值小的最优值，
那么线段树维护的是最大值即可。离散化的具体步骤，很简单，就是原数组排序然后记录序号。*/int n,k,seq[maxn],tmp[maxn];
int dp[maxn][2];///0代表以当前位为结尾的最长上升子序列，1代表以当前位为开头的最长序列
///ST结构
int st[maxn<<2];
void pushup(int rt) { st[rt] = max(st[rt<<1] , st[rt<<1|1]); }
void build(lrt)
{if(l==r){ st[rt]=0;return ; }int mid=l+r>>1;build(lson),build(rson),pushup(rt);
}
int query(lrt,int L,int R)///返回小于x的最大长度
{if(L<=l&&r<=R) return st[rt];int mid=l+r>>1,ans=0;if(L<=mid) ans=max(query(lson,L,R),ans);if(mid<R) ans=max(query(rson,L,R),ans);return ans;
}
void update(lrt,int pos,int x)
{if(l==r) {st[rt]=x;return;}int mid=l+r>>1;if(pos<=mid) update(lson,pos,x);if(mid<pos) update(rson,pos,x);pushup(rt);
}void init()
{int tp[maxn];///借用的暂时的数组memset(dp,0,sizeof(dp));memset(tp,0x7f,sizeof(tp));///数据范围要注意for(int i=1;i<=n;i++){int k=lower_bound(tp+1,tp+1+n,seq[i])-tp;dp[i][0]=k;tp[k]=seq[i];}memset(tp,0x7f,sizeof(tp));///每个数大小题目中达到10^9for(int i=n;i;i--){int k=lower_bound(tp+1,tp+1+n,-seq[i])-tp;dp[i][1]=k;tp[k]=-seq[i];}///初始化dp数组
}int main()
{int t;scanf("%d",&t);for(int ca=1;ca<=t;ca++){scanf("%d%d",&n,&k);for(int i=1;i<=n;i++) scanf("%d",&seq[i]),tmp[i]=seq[i];map<int,int> mp;int cnt=0;sort(tmp+1,tmp+n+1);///离散化，类似于树状数组常用技巧的操作for(int i=1;i<=n;i++) if(mp[tmp[i]]==0) mp[tmp[i]]=++cnt;init( );build(0,n,1);///用开头单调上升序列维护线段树///for(int i=1;i<=n;i++) cout<<dp[i][0]<<" ";puts("");///for(int i=1;i<=n;i++) cout<<dp[i][1]<<" ";puts("");int ans=0;for(int i=1;i<=n;i++){int l=i,r=i+k-1;if(r>n)break;if(l==1) ans=max(ans,dp[r+1][1]);else if(r==n) ans=max(ans,query(1,n,1,1,n));else ans=max(ans,dp[r+1][1]+query(1,n,1,1,mp[seq[r+1]]));update(1,n,1,mp[seq[i]],dp[i][0]);}printf("Case #%d: %d\n",ca,ans);}return 0;
}