1088 - Points in Segments
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Time Limit: 2 second(s) | Memory Limit: 32 MB |
Given n points (1 dimensional) and q segments, you have to find the number of points that lie in each of the segments. A point pi will lie in a segment A B if A ≤ pi ≤
B.
For example if the points are 1, 4, 6, 8, 10. And the segment is 0 to 5. Then there are 2 points that lie in the segment.
Input
Input starts with an integer T (≤ 5), denoting the number of test cases.
Each case starts with a line containing two integers n (1 ≤ n ≤ 105) and q (1 ≤ q ≤ 50000). The next line contains n space separated integers denoting the points in ascending order. All the integers
are distinct and each of them range in [0, 108].
Each of the next q lines contains two integers Ak Bk (0 ≤ Ak ≤ Bk ≤ 108) denoting a segment.
Output
For each case, print the case number in a single line. Then for each segment, print the number of points that lie in that segment.
Sample Input |
Output for Sample Input |
1 5 3 1 4 6 8 10 0 5 6 10 7 100000 |
Case 1: 2 3 2 |
Note
Dataset is huge, use faster I/O methods.
题解
这个不是二分答案的那种题了,是比较正常的那种集合区间中的那种元素查找的二分题目了。题意很简单,给n个有序的数,这些数分布在一个坐标轴上。给q次查找,询问在区间[x, y](这里的符号和代码的保持一致)中有多少个数据。
解法就是自己写一个二分函数upper_bound和lower_bound,然后直接计算区间即可。
代码示例
/**** *@author Shen *@title LightOj 1088 */ #include <iostream> #include <cstdio> #include <cmath> #include <algorithm> using namespace std; const int maxN = 100005; int a[maxN]; int t, tt; int n, q, x, y; int Bsearch_lower_bound(int x) { int l = 0, r = n - 1, mid = 0; while (l <= r) { mid = (l + r) >> 1; if (a[mid] < x) l = mid + 1; else r = mid - 1; } return l; } int Bsearch_upper_bound(int x) { int l = 0, r = n - 1, mid = 0; while (l <= r) { mid = (l + r) >> 1; if (a[mid] <= x) l = mid + 1; else r = mid - 1; } return l; } void solve() { scanf("%d%d", &n, &q); printf("Case %d:\n", ++tt); for (int i = 0; i < n; i++) scanf("%d", &a[i]); for (int i = 0; i < q; i++) { scanf("%d%d", &x, &y); int l = Bsearch_lower_bound(x); int r = Bsearch_upper_bound(y); printf("%d\n", r - l); } } int main() { scanf("%d", &t); while (t--) solve(); return 0; }
LightOJ 1088 Points in Segments 二分查找