ZOJ-3349 Special Subsequence 线段树优化DP
题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3349
题意：给定一个数列，序列A是一个满足|Ai-Ai-1| <= d的一个序列，其中Ai为给定数列中的元素，要保持相对顺序，求最长的序列A。
显然用DP来解决，f[i]表示以第i个数结尾时的最长长度，则f[i]=Max{ f[j]+1 | j<i && |num[j]-num[i]|<=d }。直接转移复杂度O(n^2)，超时。显然对于每个数num只要保存最大的f值就可以了，然后就是查找num[j]在区间[num[i]-d, num[i]+d]的最大的f[j]，用颗线段树维护就可以了。。
1 //STATUS:C++_AC_480MS_3008KB 2 #include <functional> 3 #include <algorithm> 4 #include <iostream> 5 //#include <ext/rope> 6 #include <fstream> 7 #include <sstream> 8 #include <iomanip> 9 #include <numeric> 10 #include <cstring> 11 #include <cassert> 12 #include <cstdio> 13 #include <string> 14 #include <vector> 15 #include <bitset> 16 #include <queue> 17 #include <stack> 18 #include <cmath> 19 #include <ctime> 20 #include <list> 21 #include <set> 22 #include <map> 23 using namespace std; 24 //#pragma comment(linker,"/STACK:102400000,102400000") 25 //using namespace __gnu_cxx; 26 //define 27 #define pii pair<int,int> 28 #define mem(a,b) memset(a,b,sizeof(a)) 29 #define lson l,mid,rt<<1 30 #define rson mid+1,r,rt<<1|1 31 #define PI acos(-1.0) 32 //typedef 33 typedef long long LL; 34 typedef unsigned long long ULL; 35 //const 36 const int N=100010; 37 const int INF=0x3f3f3f3f; 38 const int MOD=1e+7,STA=8000010; 39 const LL LNF=1LL<<60; 40 const double EPS=1e-8; 41 const double OO=1e15; 42 const int dx[4]={-1,0,1,0}; 43 const int dy[4]={0,1,0,-1}; 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 45 //Daily Use ... 46 inline int sign(double x){return (x>EPS)-(x<-EPS);} 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;} 50 template<class T> inline T Min(T a,T b){return a<b?a:b;} 51 template<class T> inline T Max(T a,T b){return a>b?a:b;} 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} 56 //End 57 58 int num[N],order[N],hig[N<<2],f[N]; 59 int n,d; 60 61 void update(int l,int r,int rt,int w,int val) 62 { 63 if(l==r){ 64 hig[rt]=max(hig[rt],val); 65 return ; 66 } 67 int mid=(l+r)>>1; 68 if(w<=mid)update(lson,w,val); 69 else update(rson,w,val); 70 hig[rt]=max(hig[rt<<1],hig[rt<<1|1]); 71 } 72 73 int query(int l,int r,int rt,int L,int R) 74 { 75 if(L<=l && r<=R){ 76 return hig[rt]; 77 } 78 int mid=(l+r)>>1,ret=0; 79 if(L<=mid)ret=max(ret,query(lson,L,R)); 80 if(R>mid)ret=max(ret,query(rson,L,R)); 81 return ret; 82 } 83 84 int main() 85 { 86 // freopen("in.txt","r",stdin); 87 int i,j,m,L,R,lnum,rnum,ans,w; 88 while(~scanf("%d%d",&n,&d)) 89 { 90 for(i=0;i<n;i++){ 91 scanf("%d",&num[i]); 92 order[i]=num[i]; 93 } 94 sort(order,order+n); 95 m=unique(order,order+n)-order; 96 97 mem(hig,0);ans=0; 98 for(i=0;i<n;i++){ 99 lnum=num[i]-d;rnum=num[i]+d; 100 L=lower_bound(order,order+m,lnum)-order; 101 R=upper_bound(order,order+m,rnum)-order-1; 102 f[i]=query(0,m-1,1,L,R)+1; 103 w=lower_bound(order,order+m,num[i])-order; 104 update(0,m-1,1,w,f[i]); 105 ans=max(ans,f[i]); 106 } 107 108 printf("%d\n",ans); 109 } 110 return 0; 111 }
