There a sequence S with n integers , and A is a special subsequence that satisfies |Ai-Ai-1| <= d ( 0 <i<=|A|))
Now your task is to find the longest special subsequence of a certain sequence S
Input
There are no more than 15 cases , process till the end-of-file
The first line of each case contains two integer n and d ( 1<=n<=100000 , 0<=d<=100000000) as in the description.
The second line contains exact n integers , which consist the sequnece S .Each integer is in the range [0,100000000] .There is blank between each integer.
There is a blank line between two cases
Output
For each case , print the maximum length of special subsequence you can get.
Sample Input
5 2 1 4 3 6 5 5 0 1 2 3 4 5
Sample Output
3 1
题解：让求|Ai-Ai-1| <= d 的最长子序列；
很简单就想到DP；
dp[i]表示前i个数的最长为多少，
则dp[i]=max(dp[j]+1) abs(a[i]-a[j])<=d
不出意外，肯定超时了；
代码：
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<map>
using namespace std;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
typedef long long LL;
#define mem(x,y) memset(x,y,sizeof(x))
#define PI(x) printf("%d",x)
#define PL(x) printf("%lld",x)
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%lld",&x)
#define P_ printf(" ")
#define T_T while(T--)
const int MAXN=100010;
int dp[MAXN];
int m[MAXN];
int main(){
int n,d;
while(~scanf("%d%d",&n,&d)){
for(int i=0;i<n;i++)SI(m[i]);
mem(dp,0);
int ans=0;
for(int i=0;i<n;i++){
for(int j=0;j<i;j++){
if(abs(m[i]-m[j])<=d){
dp[i]=max(dp[i],dp[j]+1);
ans=max(ans,dp[i]);
}
}
}
printf("%d
",ans+1);
}
return 0;
}
接下来就是如何高效地找到满足差值在d以内的最大值。
将数字进行离散化，对于一个新的数，就可以确定一个范围，然后在这个范围进行查找dp的最值+1即可。
线段树每一个结点保存的是这个区间的最值，叶子结点的值便是以这个数结尾的最长数量；
代码：
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<map>
using namespace std;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
typedef long long LL;
#define mem(x,y) memset(x,y,sizeof(x))
#define PI(x) printf("%d",x)
#define PL(x) printf("%lld",x)
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%lld",&x)
#define P_ printf(" ")
#define T_T while(T--)
#define ll root<<1
#define rr root<<1|1
#define lson ll,l,mid
#define rson rr,mid+1,r
#define V tree[root]
const int MAXN=100010;
int tree[MAXN<<2];
int a[MAXN],b[MAXN];
int dp[MAXN];
int nm;
int pushup(int root){
tree[root]=max(tree[ll],tree[rr]);
}
void update(int root,int l,int r,int nt,int v){
int mid=(l+r)>>1;
if(l==r){
V=v;return;
}
if(mid>=nt)update(lson,nt,v);
else update(rson,nt,v);
pushup(root);
}
void query(int root,int l,int r,int L,int R){
if(l>=L&&r<=R){
nm=max(nm,V);
return;
}
int mid=(l+r)>>1;
if(mid>=L)query(lson,L,R);
if(mid<R)query(rson,L,R);
}
int main(){
int n,d;
while(~scanf("%d%d",&n,&d)){
for(int i=0;i<n;i++)SI(a[i]),b[i]=a[i];
sort(b,b+n);
mem(tree,0);mem(dp,0);
for(int i=0;i<n;i++){
int l,r;
l=lower_bound(b,b+n,a[i]-d)-b+1;
r=upper_bound(b,b+n,a[i]+d)-b+1-1;
nm=0;
query(1,1,n,l,r);
dp[i]=nm+1;
update(1,1,n,lower_bound(b,b+n,a[i])-b+1,dp[i]);
}
int ans=0;
for(int i=0;i<n;i++)ans=max(ans,dp[i]);
printf("%d
",ans);
}
return 0;
}