POJ 2828 Buy Tickets (线段树)

题意：给定 n 个人，每次在第 pos 位置，插入一个人，然后后面的向后移动，问你最后的顺序是什么。

析：肯定不能模拟，要不然，肯定TLE，然后我们可以倒着考虑，最后一个人的位置是肯定能确定的，因为他就在最后才插入的，然后剩下的n-1个时，第n-1个人的相对位置也就确定了，依次，这样就可以用线段树来维护了，每个结点代表每个人区间已经放入多少人了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2e5 + 5;
const int mod = 10;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

int a[maxn], b[maxn];
int sum[maxn<<2];
int ans[maxn];

void push_up(int rt){ sum[rt] = sum[rt<<1] + sum[rt<<1|1]; }

void update(int M, int val, int l, int r, int rt){
  if(l == r){
    sum[rt] = 1;
    ans[l-1] = val;
    return ;
  }
  int m = l + r >> 1;
  if(M <= m-l+1-sum[rt<<1])  update(M, val, lson);
  else  update(M - (m-l+1-sum[rt<<1]), val, rson);
  push_up(rt);
}

int main(){
  while(scanf("%d", &n) == 1){
    for(int i = 0; i < n; ++i)
      scanf("%d %d", a+i, b+i);
    memset(sum, 0, sizeof sum);
    for(int i = n-1; i >= 0; --i)
      update(a[i]+1, b[i], 1, n, 1);
    for(int i = 0; i < n; ++i){
      if(i)  putchar(' ');
      printf("%d", ans[i]);
    }
    printf("\n");
  }
  return 0;
}