poj2828

题意：人们一个一个的来排队并插队，按人到来的顺序给出每个人插队的位置（插在第几个人后面），并告知每个人的id号，输出最终队伍的情况。

分析：可以用线段树，从最后来的一个人开始向来得更早的人遍历，这样pos（插在第几个人后面）的意义就变成了，前面有多少个空位。线段树上每个节点中存储的是当前时刻，该区间有多少空位。

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
usingnamespace std;

#define maxn 200005

struct Node
{
    Node *pleft, *pright;
    int num, l, r, pos;
}tree[maxn *10];

struct Item
{
    int pos, val;
}f[maxn];

int n, ncount, ans[maxn];

void input()
{
    for (int i =0; i < n; i++)
        scanf("%d%d", &f[i].pos, &f[i].val);
}

void buildtree(Node *proot, int l, int r)
{
    proot->num = r - l +1;
    proot->l = l;
    proot->r = r;
    if (l == r)
    {
        proot->pos = l;
        return;
    }
    ncount++;
    proot->pleft = tree + ncount;
    ncount++;
    proot->pright = tree + ncount;
    int mid = (l + r) /2;
    buildtree(proot->pleft, l, mid);
    buildtree(proot->pright, mid +1, r);
}

void ins(Node *proot, int num, int val)
{
    proot->num--;
    if (proot->l == proot->r)
    {
        ans[proot->pos] = val;
        return;
    }
    if (num >= proot->pleft->num)
        ins(proot->pright, num - proot->pleft->num, val);
    else
        ins(proot->pleft, num, val);
}

void work()
{
    for (int i = n -1; i >=0; i--)
        ins(tree, f[i].pos, f[i].val);
    printf("%d", ans[0]);
    for (int i =1; i < n; i++)
        printf(" %d", ans[i]);
    printf("\n");
}

int main()
{
    //freopen("D:\\t.txt", "r", stdin);
while (scanf("%d", &n) != EOF )
    {
        input();
        buildtree(tree, 0, n -1);
        work();
    }
    return0;
}