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HDU_3193_Find the hotel

2018-06-17 21:47:11来源：未知阅读 ()

Find the hotel

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 767 Accepted Submission(s): 263

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 <p_i, d_i> for a hotel h_i, where p_i stands for the price and d_i 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 h_i, 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 (p_i, d_i) 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

对于二元组(p,d)找凸点
和hdu5517类似，但是这道题的数据范围是1e5，用树状数组做是开不下来的
但是这题可以用ST表RMQ来解决
对于每一个询问(pi,di)我们找p值小于pi的二元组，看其中的最小d的值，如果di<d我们就把(pi,di)添加进最终答案
对于特殊的，如果p0是p值中的最小值，那么所有的(p0,d)组合都可以添加进最终答案
从这道题的思路来讲，其实hdu5517也可以用这道题的方法做，就不用纠结二维树状数组的问题了

 1 #include <iostream>
 2 #include <string>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <cmath>
 8 #include <vector>
 9 #include <queue>
10 #include <stack>
11 #include <set>
12 #include <map>
13 using namespace std;
14 typedef long long           LL ;
15 typedef unsigned long long ULL ;
16 const int    maxn = 1e5 + 10   ;
17 const int    inf  = 0x3f3f3f3f ;
18 const int    npos = -1         ;
19 const int    mod  = 1e9 + 7    ;
20 const int    mxx  = 100 + 5    ;
21 const double eps  = 1e-6       ;
22 const double PI   = acos(-1.0) ;
23 
24 struct node{
25     int p, d;
26 };
27 node h[maxn];
28 int cmp(const node &l, const node &r){
29     if(l.p!=r.p)
30         return l.p<r.p;
31     else
32         return l.d<r.d;
33 }
34 int Lower_Bound(int l, int r, int x){
35     while(l<r){
36         int m=(l+r)>>1;
37         if(h[m].p>=x)
38             r=m;
39         else
40             l=m+1;
41     }
42     return l;
43 }
44 int n, u, v, fac[30], st[maxn], cnt;
45 int dp[maxn][30], pl, pr, id;
46 int main(){
47     // freopen("in.txt","r",stdin);
48     // freopen("out.txt","w",stdout);
49     for(int i=0;i<30;i++)
50         fac[i]=(1<<i);
51     while(~scanf("%d",&n)){
52         for(int i=1;i<=n;i++){
53             scanf("%d %d",&h[i].p,&h[i].d);
54             dp[i][0]=i;
55         }
56         sort(h+1,h+1+n,cmp);
57         int k=(int)(log((double)n)/log(2.0));
58         for(int j=1;j<=k;j++)
59             for(int i=1;i+fac[j]-1<=n;i++){
60                 pl=dp[i][j-1];
61                 pr=dp[i+fac[j-1]][j-1];
62                 if(h[pl].d<h[pr].d)
63                     dp[i][j]=pl;
64                 else
65                     dp[i][j]=pr;
66             }
67         u=1;
68         cnt=0;
69         st[cnt++]=1;
70         for(int i=2;i<=n;i++){
71             v=Lower_Bound(1,i,h[i].p)-1;
72             if(u<=v){
73                 k=(int)(log((double)v)/log(2.0));
74                 pl=dp[u][k];
75                 pr=dp[v-fac[k]+1][k];
76                 if(h[pl].d<h[pr].d)
77                     id=pl;
78                 else
79                     id=pr;
80                 if(!(h[id].d<h[i].d))
81                     st[cnt++]=i;
82             }else{
83                 st[cnt++]=i;
84             }
85         }
86         printf("%d\n",cnt);
87         for(int i=0;i<cnt;i++)
88             printf("%d %d\n",h[st[i]].p,h[st[i]].d);
89     }
90     return 0;
91 }