HDU 5289 Assignment （线段树）

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.

题意

按序给出 n 个职工的能力值 ai 。问存在多少个 group ？其中 group 被定义为在同一个 group 中的任意两个职工的能力差不能大于等于 k ，且同一 group 中的职工编号连续。每个职工可以属于多个 group 。

解题思路

为避免重复统计，以编号 i 的职工对结果的贡献仅统计以其为 group 左区间，右区间为 R 的贡献。

对于编号 i 的职工，查找满足 aj≥ai+k 或 aj≤ai−k 的最小 j ，并将 j 作为编号 i 对结果贡献的右边界。其中 j 必须大于 i。

则对于职工 i 的真正右边界，即为区间 [i, j] 的所有职工的右边界的最小值，通过线段树求最小值。

代码

#include<bits/stdc++.h>using namespace std;const int N = 100000 + 10;int T, n, k, a[N], r[N];pair<int, int> p, q;set<pair<int, int> > st;set<pair<int, int> >::iterator it, ip;vector<set<pair<int, int> >::iterator > v;bool operator<(pair<int, int> a, pair<int, int> b) {    if(a.first == b.first)        return a.second < b.second;    return a.first < b.first;}void ERASE() {    for(int i=0;i<v.size();i++)        st.erase(v[i]);    v.clear();}const int TREE_SIZE = (N << 2) + 10;class IntervalTree{    private:        int Cover[TREE_SIZE],Top[TREE_SIZE];        int size;        int _Query(int a,int b,int l,int r,int idx)        {            if(a <= l && b >= r)    return Top[idx];            int mid = (l+r)>>1, ret = Cover[idx];            if(a<=mid)  ret = min(ret,_Query(a,b,l,mid,idx<<1));            if(b>mid)   ret = min(ret,_Query(a,b,mid+1,r,(idx<<1)+1));            return ret;        }        void _Modify(int a,int l,int r,int idx,int d)        {            if(l==r && l==a)            {                Cover[idx]=Top[idx]=d;                return;            }            int mid = (l+r)>>1;            if(a<=mid)  _Modify(a,l,mid,idx<<1,d);            else    _Modify(a,mid+1,r,(idx<<1)+1,d);            Top[idx]=min(Top[idx<<1],Top[(idx<<1)+1]);        }    public:        IntervalTree(){            memset(Cover,0x3f,sizeof(Cover));            memset(Top,0x3f,sizeof(Top));            size = (TREE_SIZE>>2) - 1;        }        IntervalTree(int size):size(size){            memset(Cover,0x3f,sizeof(Cover));            memset(Top,0x3f,sizeof(Top));        }        void init() {            memset(Cover, 0x3f, sizeof(Cover));            memset(Top, 0x3f, sizeof(Top));            size = (TREE_SIZE>>2) - 1;        }        int Query(int a,int b){            return _Query(a,b,1,size,1);        }        void Modify(int a,int d){            return _Modify(a,1,size,1,d);        }} rmq;int main(){    scanf("%d",&T);    while(T-- && scanf("%d %d",&n,&k)!=EOF)    {        rmq.init();        for(int i=1;i<=n;i++)            scanf("%d",&a[i]),  r[i] = n;        st.clear();        for(int i=1;i<=n;i++)        {            p = make_pair(a[i]-k, n+1);            it = st.upper_bound(p);            while(1) {                if(it == st.begin())    break;                it--;                r[it->second] = i-1;                v.push_back(it);            }            ERASE();            p = make_pair(a[i]+k, 0);            it = st.lower_bound(p);            for(;it!=st.end();it++) {                r[it->second] = i-1;                v.push_back(it);            }            ERASE();            st.insert(make_pair(a[i], i));        }        for(int i=1;i<=n;i++)            rmq.Modify(i, r[i]);        long long ans = 0;        for(int i=1;i<=n;i++)            ans += rmq.Query(i, r[i])-i+1;        printf("%I64d\n", ans);    }}