Buy Tickets
| Time Limit: 4000MS | Memory Limit: 65536K | |
| Total Submissions: 17416 | Accepted: 8646 |
Description
Railway tickets were difficult to buy around the Lunar New Year in China, so we must get up early and join a long queue…
The Lunar New Year was approaching, but unluckily the Little Cat still had schedules going here and there. Now, he had to travel by train to Mianyang, Sichuan Province for the winter camp selection of the national team of Olympiad in Informatics.
It was one o’clock a.m. and dark outside. Chill wind from the northwest did not scare off the people in the queue. The cold night gave the Little Cat a shiver. Why not find a problem to think about? That was none the less better than freezing to death!
People kept jumping the queue. Since it was too dark around, such moves would not be discovered even by the people adjacent to the queue-jumpers. “If every person in the queue is assigned an integral value and all the information about those who have jumped the queue and where they stand after queue-jumping is given, can I find out the final order of people in the queue?” Thought the Little Cat.
Input
There will be several test cases in the input. Each test case consists of N + 1 lines where N (1 ≤ N ≤ 200,000) is given in the first line of the test case. The next N lines contain the pairs of values Posi and Vali in the increasing order of i (1 ≤ i ≤ N). For each i, the ranges and meanings of Posi and Vali are as follows:
- Posi ∈ [0, i − 1] — The i-th person came to the queue and stood right behind the Posi-th person in the queue. The booking office was considered the 0th person and the person at the front of the queue was considered the first person in the queue.
- Vali ∈ [0, 32767] — The i-th person was assigned the value Vali.
There no blank lines between test cases. Proceed to the end of input.
Output
For each test cases, output a single line of space-separated integers which are the values of people in the order they stand in the queue.
Sample Input
4 0 77 1 51 1 33 2 69 4 0 20523 1 19243 1 3890 0 31492
Sample Output
77 33 69 51 31492 20523 3890 19243
Hint
The figure below shows how the Little Cat found out the final order of people in the queue described in the first test case of the sample input
思路：线段树。
其实就是找第K大的数，被这题坑了好多时间，本来开始咋都想不通。本来打算放弃的。
下面说一下：Posi ∈ [0, i − 1]这个条件很关键。我们从最后一个开始选取，最后一个是很容易定的，因为它就是当前所有数的第aa[i].x+1大的位置上，可以把他看成第aa[i].x+1大的数，
那么当最后一个选完后，倒数第二个就可以看成最后一个了，那么倒数第二个就是当前剩余位置中排序第aa[i-1].x+1的位置上，然后这样选取到最后就行了。其中选取的过程用线段树维护就行。复杂度N*log(n)*log(n);
1 #include<stdio.h>
2 #include<algorithm>
3 #include<iostream>
4 #include<math.h>
5 #include<stdlib.h>
6 #include<string.h>
7 using namespace std;
8 int tree[5*200000];
9 int id[200005];
10 typedef struct pp
11 {
12 int x;
13 int y;
14 } ss;
15 ss num[200005];
16 int id1[200005];
17 int ask(int l,int r,int k,int ans);
18 void build(int l,int r,int k);
19 void up(int k);
20 int main(void)
21 {
22 int i,j,k,p,q;
23 while(scanf("%d",&k)!=EOF)
24 { memset(tree,0,sizeof(tree));
25 memset(id1,0,sizeof(id1));
26 for(i=0; i<k; i++)
27 {
28 scanf("%d %d",&num[i].x,&num[i].y);
29 num[i].x+=1;
30 }
31 build(0,k-1,0);
32 for(i=k-1; i>=0; i--)
33 {
34 int er=ask(0,k-1,0,num[i].x);
35 id1[er]=num[i].y;
36 }
37 printf("%d",id1[0]);
38 for(i=1; i<k; i++)
39 printf(" %d",id1[i]);
40 printf("\n");
41 }
42 return 0;
43 }
44 void build(int l,int r,int k)
45 {
46 if(l==r)
47 {
48 tree[k]=1;
49 id[l]=k;
50 return ;
51 }
52 else
53 {
54 build(l,(l+r)/2,2*k+1);
55 build((l+r)/2+1,r,2*k+2);
56 tree[k]=tree[2*k+1]+tree[2*k+2];
57 }
58 }
59 void up(int k)
60 {
61 tree[k]=0;
62 if(k==0)return ;
63 else
64 {
65 int cc=k;
66 cc=(cc-1)/2;
67 while(cc>=0)
68 {
69 tree[cc]=tree[2*cc+1]+tree[2*cc+2];
70 if(cc==0)
71 return ;
72 cc=(cc-1)/2;
73 }
74 }
75 }
76 int ask(int l,int r,int k,int ans)
77 {
78 if(ans==tree[k])
79 {
80 int c=id[r];
81 if(tree[c]==1)
82 {
83 up(c);
84 return r;
85 }
86 else
87 {
88 if(tree[2*k+1]<ans)
89 {
90 return ask((l+r)/2+1,r,2*k+2,ans-tree[2*k+1]);
91 }
92 else if(tree[2*k+1]==ans)
93 {
94 return ask(l,(l+r)/2,2*k+1,ans);
95 }
96 }
97 }
98 else if(ans<tree[k])
99 {
100 if(tree[2*k+1]>=ans)
101 {
102 return ask(l,(l+r)/2,2*k+1,ans);
103 }
104 else if(tree[2*k+1]<ans)
105 {
106 return ask((l+r)/2+1,r,2*k+2,ans-tree[2*k+1]);
107 }
108 }
109 }