Hiho Coder-1299 Discount Ticket

description

  because Miss Shinjuku's "little Sister", The island Niang plans to go to Tokyo again in June, but this time it looks like she needs her own pocket. After a few days of battle taking, the island Niang finally in exhaustion, with Python grabbed all June, Shanghai to Tokyo, all the total n tickets. Now, please help the debt-laden island Niang to screen out the ticket for the time interval requirements. The first line of the

input

Input data contains two integers n ,? M (1?≤? N ,? m ? ≤?105), which represents the total number of tickets and the total number of queries. Next n lines, two integers per line t ,? v (1?≤? T ,? v ? ≤?105), indicating the time and price of each ticket departure. The next m line, two integers per line a ,? b (1?≤?a?≤?b?≤?105), which represents the time interval required for each query.

output

For each set of queries, the output line represents the price of . If there is no ticket that meets the requirements, the output line is "None".

Sample Input

 7 All-in-all-in-a-kind, 5 

sample Output

 919none5none 

time limit: 10000ms single point time: 1000ms memory limit: 256mb

PS. Really want to vomit groove that picture = day =

This is a static line-of-tree problem.

#include <cstdio> #include <algorithm> #include <cstring>using namespace std;const int maxn = 1e6+7;int a[maxn];struct node{int L, r, X;}    Tickt[maxn*4];void buildtree (int x, int left, int.) {TICKT[X].L = left;    TICKT[X].R = right;        if (left = = right) {tickt[x].x = A[left];    Return    } int mid = (left+right) >> 1;    Buildtree (X<<1, left, mid);    Buildtree (x<<1|1, mid+1, right); tickt[x].x = Max (tickt[x<<1].x, tickt[x<<1|1].x);} int query (int x, int left, int right) {if (left <= tickt[x].l && TICKT[X].R <= right) return tickt[    x].x;    int ans = 0;    int mid = (tickt[x].l + tickt[x].r) >> 1;    If (mid >= left) ans = max (ans, query (x<<1, left, right));    If (Mid < right) ans = max (ans, query (x<<1|1, left, right)); return ans;}    int main () {int m, n;    scanf ("%d%d", &n, &m);    memset (A, 0, sizeof (a));  for (int i = 1; I <= n; i++) {      int x, y;        scanf ("%d%d", &x, &y);    A[X] = max (a[x], y);    } buildtree (1, 1, 100000);        for (int i = 1; I <= m; i++) {int from, to;        scanf ("%d%d", &from, &to);        int p = Query (1, from, to);        if (p = = 0) printf ("none\n");    else printf ("%d\n", p); } return 0;}

Hiho Coder-1299 Discount Ticket