poj Buy Tickets（线段树）

Buy Tickets

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 7802		Accepted: 3710

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 Nlines contain the pairs of values Pos_i and Val_i in the increasing order of i (1 ≤ i ≤ N). For each i, the ranges and meanings of Pos_i and Val_i are as follows:

• Pos_i ∈ [0, i − 1] — The i-th person came to the queue and stood right behind the Pos_i-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.
• Val_i ∈ [0, 32767] — The i-th person was assigned the value Val_i.

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

这道题想了挺久的，最后实在忍不住看了别人的思路

然后发现又是一道倒着处理输入的题目，让人联想起了http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3563

这道并查集的变形题，也是倒着处理，解决了并查集没有删除的缺憾

这对以后做题也许是一个启示，但现在还没有太多感觉

回到这道题目吧：由于最后插入的人的位置一定是固定的，（因为后面不会再有人把他向后挤）

所以将这个人位置排好后，在确定下一个人的位置前把这个已排好的位置“丢掉”

这样对于插入到第i个位置，即搜索第i+1个空位

这种做法的原理并不难理解，关键在搜索函数如何设计，这个是需要自己认真独立的去思考的

我处理时将搜索和修改合并在了一起，因为每次都要搜索到叶子，然后再改回到根

/*自己做的时候还有出了个问题，由于看到需要先输入，再倒着处理，就用了栈

结果超时了，改回数组就没问题了，切记以后能不用stl就不要用，效率要低一些*/

代码如下:

#include <stdio.h>

int p[200005];
int v[200005];
int ans[200005];
struct Node{
    int l,r,num;
}tree[1000000];

void Build(int n,int x,int y){
    tree[n].l = x;
    tree[n].r = y;
    tree[n].num = (y - x + 1);//储存的是该区间内的空位数
    int mid = (x + y) / 2;
    if(x == y){
        return;
    }
    Build(2*n,x,mid);
    Build(2*n+1,mid+1,y);
}

void Modify(int n,int p,int v){
    int l = tree[n].l;
    int r = tree[n].r;
    int mid = (l + r) / 2;
    if(l == r){
        tree[n].num = 0;
        ans[l] = v;
        return;
    }
    if(tree[2*n].num >= p)  Modify(2*n,p,v);
    else                    Modify(2*n+1,p - tree[2*n].num,v);
    tree[n].num = tree[2*n].num + tree[2*n+1].num;
}

int main(){
    int n;
    int i;

    while(scanf("%d",&n) != EOF){
        Build(1,1,n);
        for(i = 0;i < n;i++){
            scanf("%d%d",&p[i],&v[i]);
        }
        for(i = n - 1;i >= 0;i--){
            Modify(1,p[i]+1,v[i]);
        }
        for(i = 1;i < n;i++){
            printf("%d ",ans[i]);
        }
        printf("%d\n",ans[n]);
    }

    return 0;
}