HDU 5089 Assignment (rmq + binary or monotonic Queue)

HDU 5089 Assignment (rmq + binary or monotonic Queue)

 

Assignment Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission (s): 557 Accepted Submission (s): 280


Problem Description Tom owns a company and he is the boss. there are n staffs which are numbered from 1 to n in this company, and every staff has a ability. now, Tom is going to assign a special task to some staffs who were in the same group. in a group, the difference of the ability of any two staff is less than k, and their numbers are continuous. tom want to know the number of groups like this.
Input In the first line a number T indicates the number of test cases. Then for each case the first line contain 2 numbers n, k (1 <= n <= 100000, 0 Output For each test, output the number of groups.
Sample Input

24 23 1 2 410 50 3 4 5 2 1 6 7 8 9

Sample Output

528HintFirst Sample, the satisfied groups include:[1,1]、[2,2]、[3,3]、[4,4] 、[2,3]  

Author FZUACM
Source 2015 Multi-University Training Contest 1
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Rmq + binary

 

#include
  
   #include
   
    #include
    
     #include#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          #define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8typedef __int64 ll;const int mod=1e9+7;using namespace std;#define N 100007int a[N],n;int dpmin[N][25],dpmax[N][25];int k;inline bool judge(int le,int ri){ int kk=log2((ri-le+1)*1.0); int mi=min(dpmin[le][kk],dpmin[ri-(1<
           
            >1; if(judge(i,mid)) { p=mid; le=mid+1; } else ri=mid-1; } ans+=p-i+1; } printf(%I64d,ans); } return 0;}
           
         
        
       
      
     
    
   
  

 

 

Monotonous queue

#include
  
   #include
   
    #include
    
     #include#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          #define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8using namespace std;const int mod=1e9+7;#define INF 0x3f3f3f3fconst int N=100005;int mique[N],maque[N],mihead,mahead,mitail,matail;int n,a[N],k;int pre,now;__int64 ans;inline void miinque(int i){ while(mihead
           
            =1) ans-=(__int64)(prelen+1)*(prelen)/2; pre=n-1; } printf(%I64d,ans+n); } return 0;}