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HDU 5089 Assignment（rmq+二分或單調隊列）

編輯：關於C++

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

2
4 2
3 1 2 4
10 5
0 3 4 5 2 1 6 7 8 9

Sample Output

5
28
HintFirst 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+二分

#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-8
typedef __int64 ll;
const int mod=1e9+7;

using namespace std;

#define N 100007

int 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;
}

單調隊列

#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-8

using namespace std;
const int mod=1e9+7;

#define INF 0x3f3f3f3f

const 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;
}