POJ 2828 Buy Tickets (線段樹)

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Buy Tickets

Time Limit: 4000MS		Memory Limit: 65536K
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 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

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, Zeyuan 題目的意思就是給定N個插隊的順序，求最後隊伍的排序看了下題目給的資料20W，鏈表（定址）和數組（數組移動）會逾時。看了題解才知道用線段樹。線段樹可以精確尋找到位置。讀入插隊順序的時候如果從前往後讀的話，每次插入後，有的人的當前位置會改變。所以採用從後往前讀的方法因為因為最後一個人的位置是固定的，所以從後往前排，每次排的那個人的位置就固定了。拿範例分析一下：第一次在第0位插：77第二次在第一位插51：77 51第三次在第一位插33：77 33 51第四次在第二位插69：77 33 69 51 逆序的話：第一次在前面還有2個空位的地方插69：_ _ 69 _第二次在前面還有1個空位的地方插33：_ 33 69 _第三次在前面還有1個空位的地方插51：_ 33 69 51第四次在前面還有0個空位的地方插77:77 33 69 51不過這道題目的想法很重要，每個節點表示這個區間內的空的位置的數量，從後往前讀，pos的值表示這個人前面有多少個空位，然後在樹裡面尋找空位。

 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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代碼跑起來有三千七百多秒，還要再最佳化一下。