AssignmentTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1565 Accepted Submission(s): 754 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 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 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] Author FZUACM Source 2015 Multi-University Training Contest 1 Recommend We have carefully selected several similar problems for you: 5309 5308 5307 5306 5305 |
题解： 以数组d[i]表示区间以a[i]结尾满足任意两元素差值小于k的最大长度，
则使用线段树查询在a[i]之前与其差值大于等于k的位置为p
d[i] = max(d[i-1], p+1);
答案即为 sum = 西格玛（d[i] - i + 1）;
题目很简单，关键是思路要精确 ， 很多时候如果数学思维卡住了，可以使用编程思维来进行解决. 还是很喜欢使用线段树进行解题，多加努力。
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
using namespace std;
#define maxn 100000 + 10
#define lson L, mid, rt<<1
#define rson mid+1, R, rt<<1|1
int n, k;
int a[maxn];
int ans;
struct Node
{
int mi, ma;
}T[maxn<<2];
void pushup(int rt)
{
int l = rt<<1, r = rt<<1|1;
T[rt].ma = max(T[l].ma, T[r].ma);
T[rt].mi = min(T[l].mi, T[r].mi);
}
void build(int L, int R, int rt)
{
if(L == R)
{
T[rt].ma = T[rt].mi = a[L];
return ;
}
int mid = (L + R) >> 1;
build(lson);
build(rson);
pushup(rt);
}
///注意树的节点与区间节点不要混淆
void query(int l, int r, int v, int L, int R, int rt)
{
if(L == R)
{
if(abs(T[rt].mi - v) >= k)
{
if(ans == -1 || ans < L)
ans = L;
}
return ;
}
int mid = (L + R) >> 1;
if(r > mid)
if(abs(T[rt<<1|1].mi - v) >= k || abs(T[rt<<1|1].ma - v) >= k)
query(l, r, v, rson);
if(ans == -1 && l <= mid)
if(abs(T[rt<<1].mi - v) >= k || abs(T[rt<<1].ma - v) >= k)
query(l, r, v, lson);
}
int d[maxn];
long long sum;
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
sum = 1;
scanf("%d%d", &n, &k);
for(int i=1; i<=n; i++)
scanf("%d", &a[i]);
build(1, n, 1);
d[1] = 1;
for(int i=2; i<=n; i++)
{
ans = -1;
query(1, i, a[i], 1, n, 1);
if(ans == -1)
d[i] = d[i-1];
else d[i] = max(d[i-1], ans + 1);
sum += i - d[i] + 1;
}
printf("%I64d\n", sum);
}
return 0;
}
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