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.
Source
POJ Monthly--2006.05.28, Zhu, Zeyuan
題目大意，插隊的問題，每個案例給出n，代表有n個插隊的，每個給出p,v，意思是代號為v的人插在了第p個人的後面，問最後的隊伍的排列？
題目中一開始整個隊列是空的，也就是說如果輸入i，j：代表代號為j的人插在了第i個人的後面，也就是說在他之前一定有了i個人，而他的位置是i+1。
所以由後向前推到，每個遇到的都是確定位置的，最後的人選定的位置不會改變，同樣因為是倒敘輸入，在第i個人後插隊，也就是說他的前面一定要留下i個空格。
經過這樣的轉化後就可以用線段樹解決，lazy中存了代號，cl數組中存了每個節點中包含的空格的數目。
#include#include #include using namespace std; #define maxn 210000 #define lmin 1 #define rmax n #define root lmin,rmax,1 #define lson l,(l+r)/2,rt<<1 #define rson (l+r)/2+1,r,rt<<1|1 #define now l,r,rt #define int_now int l,int r,int rt struct node { int a , b ; } p[maxn]; int cl[maxn<<2] , lazy[maxn<<2] ; int n , i ; void push_up(int_now) { if( l != r ) cl[rt] = cl[rt<<1] + cl[rt<<1|1] ; } void creat(int_now) { cl[rt] = lazy[rt] = 0 ; if( l != r ) { creat(lson); creat(rson); push_up(now); } else cl[rt] = 1 ; return ; } void update(int a,int k,int_now) { if( l == r && cl[rt] == 1 ) { cl[rt] = 0 ; lazy[rt] = k ; } else { if( a <= cl[rt<<1] ) update(a,k,lson); else update(a-cl[rt<<1],k,rson); push_up(now); } return ; } void pre(int_now) { if( l == r ) { if(n == 1) printf("%d\n", lazy[rt]); else printf("%d ", lazy[rt]); n-- ; } else { pre(lson); pre(rson); } return ; } int main() { while( scanf("%d", &n)!=EOF ) { for(i = 0 ; i < n ; i++) { scanf("%d %d", &p[i].a, &p[i].b); p[i].a++ ; } creat(root); for(i = n - 1 ; i >= 0 ; i--) { update(p[i].a,p[i].b,root); } pre(root); } return 0; }
