HDU Find the hotel(RMQ)

Mr_Treeeee發表於2020-04-06
MQ

Find the hotel

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 611    Accepted Submission(s): 201


Problem Description
  Summer again! Flynn is ready for another tour around. Since the tour would take three or more days, it is important to find a hotel that meets for a reasonable price and gets as near as possible! 
  But there are so many of them! Flynn gets tired to look for any. It’s your time now! Given the <pi, di> for a hotel hi, where pi stands for the price and di is the distance from the destination of this tour, you are going to find those hotels, that either with a lower price or lower distance. Consider hotel h1, if there is a hotel hi, with both lower price and lower distance, we would discard h1. To be more specific, you are going to find those hotels, where no other has both lower price and distance than it. And the comparison is strict.
 

Input
There are some cases. Process to the end of file.
Each case begin with N (1 <= N <= 10000), the number of the hotel.
The next N line gives the (pi, di) for the i-th hotel.
The number will be non-negative and less than 10000.
 

Output
First, output the number of the hotel you find, and from the next line, print them like the input( two numbers in one line). You should order them ascending first by price and break the same price by distance.
 

Sample Input
3
15 10
10 15
8 9





 



Sample Output




1
8 9





 



Author



ZSTU



 



Source



HDU 2009-12 Programming Contest 



 


題意：


找特定旅館的數量，這個旅館的p和d不能都比其他旅館大。可以相等。





POINT：


先給p從小到大排個序，然後選定一個旅館p0 d0，從價格0-(p0-1)的範圍內找到d最小的，和d0比較，符不符合要求就是了。





#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define LL long long
const int N = 10000+5;
struct node
{
    int p,d;
}h[N],ans[N];
int dp[N][20];
int n;
bool cmd(node a,node b)
{
    if(a.p!=b.p)
    {
        return a.p<b.p;
    }
    else
        return a.d<b.d;
}
void prermq()
{
    for(int i=1;i<=n;i++)
    {
        dp[i][0]=h[i].d;
    }
    for(int j=1;1<<j<=n;j++)
    {
        for(int i=1;i-1+(1<<j)<=n;i++)
        {
            dp[i][j]=min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
        }
    }
}
int query(int l,int r)
{
    int k=(int)log2((double)(r-l+1));
    return min(dp[l][k],dp[r+1-(1<<k)][k]);
}
int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d %d",&h[i].p,&h[i].d);
        }
        sort(h+1,h+1+n,cmd);
        prermq();
        int cnt=0;
        for(int i=1;i<=n;i++)
        {
            int j=i-1;
            while(j!=0&&h[j].p==h[i].p)
            {
                j--;
            }
            if(j==0)
                ans[++cnt]=h[i];
            else
            {
                int d=query(1,j);
                if(h[i].d<=d)
                {
                    ans[++cnt]=h[i];
                }
            }
        }
        printf("%d\n",cnt);
        sort(ans+1,ans+1+cnt,cmd);
        for(int i=1;i<=cnt;i++)
        {
            printf("%d %d\n",ans[i].p,ans[i].d);
        }
    }
}








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