hdu 5289 ST表+尺取法

最新推荐文章于 2021-08-03 23:17:08 发布

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本文解析了一道关于员工能力分组的算法题，通过区间最小最大值查询的方法，求解满足条件的连续子数组数量。该问题适用于区间查询算法的学习。

题目：

Assignment

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4264 Accepted Submission(s): 1958

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

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

Sample Output

5
28
Hint
First Sample, the satisfied groups include:[1,1]、[2,2]、[3,3]、[4,4] 、[2,3]

分析：

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=100050;
const int inf=INT_MAX;

int t,n,k,a[maxn],st[maxn][16],sti[maxn][16];

inline void initrmq(){
    for(int j=1;(1<<j)<=n;++j){
        for(int i=0;i+(1<<j)-1<n;++i){
            st[i][j]=min(st[i][j-1],st[i+(1<<(j-1))][j-1]);//+比<<优先级高
            sti[i][j]=max(sti[i][j-1],sti[i+(1<<(j-1))][j-1]);
        }
    }
}

inline int rmq(int u,int v){
    int k=(int)(log(v-u+1.0)/log(2.0));
    return max(sti[u][k],sti[v-(1<<k)+1][k])-min(st[u][k],st[v-(1<<k)+1][k]);
}

/*inline void In(int &x){
    char c;x=0;c=getchar();
    int sign=1;
    while(!(c>='0'&&c<='9'||c=='-')) c=getchar();
    if(c=='-') sign=-1,c=getchar();
    while(c>='0'&&c<='9'){
        x=(x<<3)+(x<<1)+c-'0';
        c=getchar();
    }
    x*=sign;
}

inline void Out(__int64 x){
    if(x<0){
        x=-x;
        putchar('-');
    }
    if(x>9) Out(x/10);
    putchar(x%10+'0');
}*/

int main(){///
    __int64 l,r;
    __int64 ans;
    //In(t);
    scanf("%d",&t);
    while(t--){
        //In(n); In(k);
        scanf("%d%d",&n,&k);
        for(int i=0;i<n;++i){
            //In(a[i]);
            scanf("%d",&a[i]);
            st[i][0]=sti[i][0]=a[i];
        }
        initrmq();
        ans=0;
        l=0;///枚举右端点 600ms
        for(int i=0;i<n;++i){
            r=i;
            while(l<r&&rmq(l,r)>=k) ++l;///找到最左的符合条件的左端点
            ans+=(r-l+1);
        }

        /*r=0;///枚举左端点迷之T的写法.....
        for(int i=0;i<n;++i){
            l=i;
            while(r<n)
                while(rmq(l,r)<k) ++r;
            ans+=(r-l);
            //cout<<l<<"  "<<r<<endl;
        }*/
        /*枚举左端点RE的姿势：
        l=0; r=0;
        while(1){
            while(r<n&&rmq(l,r)<k) r++;
            if(r==n) break;
            ans+=r-l;
            l++;
        }
        for(;l<n;++l) ans+=n-l;*/
        printf("%I64d\n",ans);
        //Out(ans);
        //putchar('\n');
    }
    return 0;
}

/*//一个只需要300ms的姿势...
#include<stdio.h>
#include<stdlib.h>

int num[100050],n,k;
__int64 sum;

int main(){
	int i,j,T;
	scanf("%d",&T);
	while(T--){
		scanf("%d%d",&n,&k);
		for(i=0;i<n;i++){
			scanf("%d",&num[i]);
		}
		sum=n;
		int max,min,l;
		max=min=num[0];
		l=0;
		for(i=1;i<n;i++){
			if(num[i]>max)max=num[i];
			else if(num[i]<min)min=num[i];
			if(max-min<k){
				l++;
				sum+=l;
			}
			else{
				min=max=num[i];
				l=0;
				for(j=i-1;abs(num[j]-max)<k&&abs(num[j]-min)<k;j--){
					if(num[j]>max)max=num[j];
					else if(num[j]<min)min=num[j];
					l++;
				}
				sum+=l;
			}
		}
		printf("%I64d\n",sum);
	}
	return 0;
}*/