hdu 5289 关于线段树的解法 很有意思

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Assignment Time 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 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 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 
#include 
#include 
#include 

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;
}