Poj 2828 buy tickets (line segment tree)

Buy tickets

Time limit:4000 Ms		Memory limit:65536 K
Total submissions:13017		Accepted:6449

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 Olympus in ICS ICs.

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 wocould 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 consistsN+ 1 lines whereN(1 ≤N≤ 200,000) is given in the first line of the test case. The nextNLines contain the pairs of ValuesPosiAndValiIn the increasing orderI(1 ≤I≤N). For eachI, The ranges and meaningsPosiAndValiAre as follows:

• Posiε [0,I−1]-I-Th person came to the queue and stood right behindPosi-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.
• Valiε [0, 32767]-I-Th person was assigned the valueVali.

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

40 771 511 332 6940 205231 192431 38900 31492

Sample output

77 33 69 5131492 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.

Source

Poj monthly -- 2006.05.28, Zhu, and zeyuan indicate that the order of N queues is given, and the sorting of the last team is obtained. The data given by the next question is 20 W. Linked List (addressing) and array (array movement) Will time out. After reading the problem, you can see that the line segment tree is used. The line segment tree can precisely locate the position. If you read data from the forward and backward when reading data into the queue, the current position of some people will change after each insertion. So the method of reading from the back to the front is adopted because the last person's position is fixed, so the position of the person in each row is fixed from the back to the front. Take the example for analysis: the first time In The 0th-Bit Insertion: 77 the second time in the first place in the 51: 77 51 the third time in the first place in the 33: 77 33 51 the fourth time in the second place in the 69: 77 33 69 51 in reverse order: for the first time, insert 69: _ 69 _ in the first place where there are two vacant spaces. For the second time, insert 33 in the front where there is one vacant space: _ 33 69 _ insert 51: _ 33 69 51 in the front of a vacant space for the third time. Insert 77 in the front of a vacant space for the fourth time: 77 33 69 51 but the idea of this question is very important. Each node represents the number of null positions in this range and reads from the back to the front, the POS value indicates the number of vacant spaces in front of the person, and then finds the vacant space in the tree.

 1 #include<cstdio> 2 #include<cstring> 3 #include<stdlib.h> 4 #include<algorithm> 5 using namespace std; 6 const int MAXN=200000+5; 7 struct node 8 { 9     int pos;10     int val;11     int l,r;12     int mid()13     {14         return (l+r)/2;15     }16 }a[MAXN*4];17 18 struct input19 {20     int pos;21     int val;22 }num[MAXN];23 24 int ans[MAXN],cnt=1;25 26 void btree(int l,int r,int step)27 {28     a[step].l=l;29     a[step].r=r;30     if(l==r)31     {32         a[step].pos=1;33         return ;34     }35     int mid=a[step].mid();36     btree(l,mid,step*2);37     btree(mid+1,r,step*2+1);38     a[step].pos=a[step*2].pos+a[step*2+1].pos;39 }40 41 void ptree(int step,int pos,int val)42 {43     if(a[step].l==a[step].r)44     {45         a[step].val=val;46         a[step].pos=0;47         return ;48     }49     if(a[step*2].pos>pos)50         ptree(step*2,pos,val);51     else52         ptree(step*2+1,pos-a[step*2].pos,val);53     a[step].pos=a[step*2].pos+a[step*2+1].pos;54 }55 56 void fintree(int step)57 {58     if(a[step].l==a[step].r)59     {60         ans[cnt++]=a[step].val;61         return ;62     }63     fintree(step*2);64     fintree(step*2+1);65 }66 67 int main()68 {69     int n;70     while(scanf("%d",&n)!=EOF)71     {72         for(int i=1;i<=n;i++)73             scanf("%d %d",&num[i].pos,&num[i].val);74         btree(1,n,1);75         for(int i=n;i>=1;i--)76             ptree(1,num[i].pos,num[i].val);77         cnt=1;78         fintree(1);79         for(int i=1;i<n;i++)80             printf("%d ",ans[i]);81         printf("%d\n",ans[n]);82     }83     return 0;84 }

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The Code has been running for more than three thousand seven hundred seconds and needs to be optimized.