TOJ Open Ural FU Championship 2013 G

G. Nested Segments

Time limit: 1.0 second
Memory limit: 64 MB

You are given n segments on a straight line. For each pair of segments it is known that they either have no common points or all points of one segment belong to the second segment.

Then m queries follow. Each query represents a point on the line. For each query, your task is to find the segment of the minimum length, to which this point belongs.

Input

The first line contains an integer n that is the number of segments (1 ≤ n ≤ 10⁵). i’th of the next n lines contains integers a_i and b_i that are the coordinates of endpoints of the i’th segment (1 ≤ a_i < b_i ≤ 10⁹). The segments are ordered by non-decreasing a_i, and when a_i = a_j they are ordered by decreasing length. The next line contains an integer m that is the number of queries (1 ≤ m ≤ 10⁵). j’th of the next m lines contains an integer c_j that is the coordinate of the point (1 ≤ c_j ≤ 10⁹). The queries are ordered by non-decreasing c_j.

Output

For each query output the number of the corresponding segment on a single line. If the point does not belong to any segment, output “-1”. The segments are numbered from 1 to n in order they are given in the input.

Sample

input	output
3 2 10 2 3 5 7 11 1 2 3 4 5 6 7 8 9 10 11	-1 2 2 1 3 3 3 1 1 1 -1

Problem Author: Mikhail Rubinchik, idea by Nikita Pervukhin

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <queue>
 4 #include <algorithm>
 5 using namespace std;
 6 #define maxn 100055
 7 #define ll long long
 8 #define INF 0x7fffffff
 9 //#define ll __int64
10 int n, m;
11 struct T{
12     int x, cover, id, a, b, c;
13 };
14 int cmp(T n1, T n2){
15     if (n1.x != n2.x)return n1.x<n2.x;
16     return n1.c > n2.c;
17 }
18 int g[maxn * 2];
19 T f[maxn * 4];
20 T d[maxn * 4];
21 int p[maxn * 2], ans[maxn * 2];
22 int main(){
23     while (~scanf("%d", &n)){
24         memset(f, 0, sizeof f);
25         memset(p, 0, sizeof p);
26         memset(ans, 0, sizeof ans);
27         memset(d, 0, sizeof d);
28         memset(g, 0, sizeof g);
29         for (int i = 1; i <= n * 2; i += 2){
30             scanf("%d%d", &d[i].x, &d[i + 1].x);
31             d[i].cover = 1; d[i + 1].cover = -1;
32             d[i].c = d[i + 1].x - d[i].x;
33             d[i].id = (i + 1) / 2;
34         }
35         d[0].x = 0; d[2 * n + 1].x = INF; d[0].id = -1;
36         d[0].cover = 1; d[2 * n + 1].cover = -1;
37         sort(d, d + n + n + 2, cmp);
38         int cov = 0, t = -2, tc, tid = -1;
39         for (int i = 0; i < 2 * n + 1; i++){
40             if (d[i].cover == -1)f[i].a = d[i].x + 1;
41             else f[i].a = d[i].x;
42             cov += d[i].cover;
43             if (d[i].cover == 1)g[cov] = d[i].id;
44             if (d[i + 1].cover == 1)f[i].b = d[i + 1].x - 1;
45             else f[i].b = d[i + 1].x;
46             f[i].id = g[cov];
47         }
48         for (int i = 0; i < 2 * n + 1; i++){
49             p[i] = f[i].a;
50             ans[i] = f[i].id;
51         }
52         sort(f, f + 2 * n + 2, cmp);
53         int x;
54         scanf("%d", &m);
55         for (int j = 1; j <= m; j++){
56             scanf("%d", &x);
57             printf("%d\n", ans[upper_bound(p, p + 2 * n + 1, x) - p - 1]);
58         }
59     }
60     return 0;
61 }

View Code 2013-10-20 17:39:40

posted @ 2013-10-19 22:11 HaibaraAi 阅读(93) 评论(0) 收藏举报

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