搜索 - hdu5289 Assignment

题目：

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

题意：

给定一个序列，及界限值k，问该序列中存在多少个区间，满足区间内任意两个数字之差的绝对值小于k

思路：

利用RMQ算法，用nlogn的时间复杂度先算出每个区间的最大最小值

枚举每个区间的左值L，二分搜索最大的区间右值R

则最大区间[L,R]的贡献为R-L+1

代码：

#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<functional>
#pragma comment(linker, "/STACK:102400000,102400000")//C++
using namespace std;

const double PI = 3.141592653589793238462643383279502884197169399;
const int MAXINT = 0x7fffffff;
const int MAXSIZE = 100000 + 5;

int arr[MAXSIZE];
int dmin[MAXSIZE][32],dmax[MAXSIZE][32];

void RMQ_init(int A[],int len){
    //int n=A.size();
    for (int i=0;i<len;++i){
        dmin[i][0] = A[i];
        dmax[i][0] = A[i];
    }
    for (int j=1;(1<<j)<=len;++j)
        for (int i=0;i+(1<<j)-1<len;++i){
            dmin[i][j] = min(dmin[i][j-1],dmin[i+(1<<(j-1))][j-1]);
            dmax[i][j] = max(dmax[i][j-1],dmax[i+(1<<(j-1))][j-1]);
        }
    return ;
}

int RMQ_min(int L, int R){
    int k=0;
    while (1<<(k+1) <=  R-L+1) k++;
    return min(dmin[L][k],dmin[R-(1<<k)+1][k]);
}

int RMQ_max(int L, int R){
    int k=0;
    while (1<<(k+1) <=  R-L+1) k++;
    return max(dmax[L][k],dmax[R-(1<<k)+1][k]);
}

bool RMQ_check(int L, int R, int k){
    if (RMQ_max(L,R)-RMQ_min(L,R) < k)
        return true;
    else return false;
}

int main(){
    int total;
    cin>>total;
    while (total--){
        memset(arr,0,sizeof(arr));
        memset(dmin,0,sizeof(dmin));
        memset(dmax,0,sizeof(dmax));

        int n,k;
        scanf("%d %d",&n,&k);
        for (int i=0;i<n;++i) scanf("%d",arr+i);

        RMQ_init(arr,n);
        //cout<<RMQ_min(0,3)<<endl;

        long long ans=0;
        for (int i=0;i<n;++i){
            int l = i, r = n-1;
            while (l<=r){
                //cout<<"i&r&l: "<<i<<" "<<l<<" "<<r<<endl;
                int mid = (l+r)>>1;
                if (RMQ_check(i,mid,k)){
                    l = mid+1;
                }
                else r = mid-1;
            }
            //cout<<"l&r: "<<i<<" "<<l-1<<endl;
            ans+=l-i;
        }
        cout<<ans<<endl;
    }
    return 0;
}