hdu5289 Assignment（多校第一场第二题：RMQ+找规律或单调队列+找规律）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=5289

Assignment

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1770    Accepted Submission(s): 854

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<k<=10^9),indicate the company has n persons, k means the maximum difference between abilities of staff in a group is less than k. The second line contains n integers:a[1],a[2],…,a[n](0<=a[i]<=10^9),indicate the i-th staff’s ability.

 

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

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

 

Author

FZUACM

 

Source

2015 Multi-University Training Contest 1

 

AC code1（RMQ+找规律）：

#include<iostream>#include<algorithm>#include<cstring>#include<cmath>#include<cstdio>#include<queue>#include<map>#include<vector>#define LL long long#define MAXN 1000010using namespace std;int num[MAXN];int ma[100001][17],mi[100001][17];int n,k,i,j;LL ans=0;//状态转移方程ma[j][i]=max(ma[j][i-1],ma[j+(1<<(i-1))][i-1])//mi[j][i]=min(mi[j][i-1],mi[j+(1<<(i-1))][i-1])][i-1] void rmqinit() //j为区间起点，(1<<i)为区间长度 {for(j=1;j<=n;j++){ma[j][0]=mi[j][0]=num[j];}for(i=1;(1<<i)<=n;i++)//枚举区间长度为(1<<i) {for(j=1;j+(1<<i)-1<=n;j++)//枚举区间起点 {ma[j][i]=max(ma[j][i-1],ma[j+(1<<(i-1))][i-1]);mi[j][i]=min(mi[j][i-1],mi[j+(1<<(i-1))][i-1]); }}}int rmqmax(int l,int r){int k=(int)(log(1.0*(r-l+1))/log(2.0));return max(ma[l][k],ma[r-(1<<k)+1][k]);}int rmqmin(int l,int r){int k=(int)(log(1.0*(r-l+1))/log(2.0));return min(mi[l][k],mi[r-(1<<k)+1][k]);}int main(){//printf("%I64d\n",1<<17);int T;while(scanf("%d",&T)!=EOF){while(T--){scanf("%d%d",&n,&k);for(i=1;i<=n;i++){scanf("%d",&num[i]);}rmqinit();int st=1;int ed=1;ans=0;for(ed=1;ed<=n;ed++){   //下面不能用if语句判断，必须用while,因为起点st对应的元素不一定就是当前区间最大值或最小值，  //且可以肯定的是，当前ed对应的元素一定是区间最大值或最小值，  //而区间最大与最小值元素之间的元素数可能不止一个 while(rmqmax(st,ed)-rmqmin(st,ed)>=k&&st<ed){ans+=ed-st;st++;}}while(st<=n){ans+=(ed-st);st++;}printf("%I64d\n",ans);}}return 0; } 

AC code2（单调队列+找规律）：

#include<iostream>#include<algorithm>#include<cstring>#include<cmath>#include<cstdio>#include<queue>#include<map>#include<vector>#define LL long long#define MAXN 1000010using namespace std;deque<int>deq1,deq2;//deq1为单调递增队列，deq2为单调递减队列int num[MAXN];int i,j,n,k;LL ans=0;int main(){int T;scanf("%d",&T);while(T--){scanf("%d%d",&n,&k);for(i=1;i<=n;i++){scanf("%d",&num[i]);}while(!deq1.empty()) deq1.pop_back();while(!deq2.empty()) deq2.pop_back();ans=0;int st=1;int ed=1;for(ed=1;ed<=n;ed++){while(!deq1.empty()&&deq1.back()>num[ed]) deq1.pop_back();deq1.push_back(num[ed]);while(!deq2.empty()&&deq2.back()<num[ed]) deq2.pop_back();deq2.push_back(num[ed]);while(!deq1.empty()&&!deq2.empty()&&deq2.front()-deq1.front()>=k){ans+=(ed-st);if(deq1.front()==num[st]) deq1.pop_front();if(deq2.front()==num[st]) deq2.pop_front();st++;}}while(st<=n){ans+=(ed-st);st++;}printf("%I64d\n",ans);}return 0; }