Poj 2828buy tickets

Poj 2828

The question is that there are n insert operations. Each time B is inserted to position a, all of the original values after a are moved 1, and the final sequence of the ball is

In the tree, data can be inserted by scanning multiple locations in the entire interval. Therefore, data can be inserted from the scanning forward to the line segment tree.

For example, if B is inserted at location a on the left side of the entire range and data can be inserted at location a on the left side of the entire range, you can only put the data to the right of the entire range, and recursively continue.

1 void Update (int K, int L, int R, int X) 2 {3 if (L = r) {pre [k] = R; ans [l] = val; return;} 4 5 Int mid = (L + r)> 1; 6 7 if (x <= pre [k <1]) update (lson, x); // redundant x 8 9 else update on the left (rson, X-pre [k <1]); // otherwise, pre [k] = pre [k <1] + pre [k <1 | 1]; 12}

 

 1 #include <map> 2 #include <set> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <vector> 8 #include <cstdio> 9 #include <cctype>10 #include <cstring>11 #include <cstdlib>12 #include <iostream>13 #include <algorithm>14 using namespace std;15 #define INF 1e916 #define inf (-((LL)1<<40))17 #define lson k<<1, L, mid18 #define rson k<<1|1, mid+1, R19 #define mem0(a) memset(a,0,sizeof(a))20 #define mem1(a) memset(a,-1,sizeof(a))21 #define mem(a, b) memset(a, b, sizeof(a))22 #define FOPENIN(IN) freopen(IN, "r", stdin)23 #define FOPENOUT(OUT) freopen(OUT, "w", stdout)24 template<class T> T CMP_MIN(T a, T b) { return a < b; }25 template<class T> T CMP_MAX(T a, T b) { return a > b; }26 template<class T> T MAX(T a, T b) { return a > b ? a : b; }27 template<class T> T MIN(T a, T b) { return a < b ? a : b; }28 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }29 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }30 31 typedef __int64 LL;32 //typedef long long LL;33 const int MAXN = 200005;34 const int MAXM = 100005;35 const double eps = 1e-10;36 const LL MOD = 1000000007;37 38 int N, pre[MAXN<<2];39 int id[MAXN], num[MAXN], ans[MAXN];40 int r, val;41 42 int buildTree(int k, int L, int R)43 {44     if(L == R) return pre[k] = 1;45     int mid = (L+R)>>1;46     return pre[k] = buildTree(lson) + buildTree(rson);47 }48 49 void update(int k, int L, int R, int x)50 {51     if(L == R) { pre[k] = r; ans[L] = val; return ; }52 53     int mid = (L+R)>>1;54 55     if(x <= pre[k<<1]) update(lson, x);56 57     else update(rson, x-pre[k<<1]);58 59      pre[k] = pre[k<<1] + pre[k<<1|1];60 }61 62 int main()63 {64     while(~scanf("%d", &N))65     {66         buildTree(1, 1, N);67         for(int l=1;l<=N;l++)68             scanf("%d %d", &id[l], &num[l]);69         r = 0;70         for(int i=N;i>0;i--)71         {72             val = num[i];73             update(1, 1, N, id[i] + 1);74         }75         for(int i=1;i<=N;i++) printf("%d%c", ans[i], i==N?‘\n‘:‘ ‘);76     }77     return 0;78 }