A. Rudolph and Cut the Rope

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n� nails driven into the wall, the i�-th nail is driven ai�� meters above the ground, one end of the bi�� meters long rope is tied to it. All nails hang at different heights one above the other. One candy is tied to all ropes at once. Candy is tied to end of a rope that is not tied to a nail.

To take the candy, you need to lower it to the ground. To do this, Rudolph can cut some ropes, one at a time. Help Rudolph find the minimum number of ropes that must be cut to get the candy.

The figure shows an example of the first test:

Input

The first line contains one integer t� (1≤t≤1041≤�≤104) --- the number of test cases.

The first line of each test case contains one integer n� (1≤n≤501≤�≤50) --- the number of nails.

The i�-th of the next n� lines contains two integers ai�� and bi�� (1≤ai,bi≤2001≤��,��≤200) --- the height of the i�-th nail and the length of the rope tied to it, all ai�� are different.

It is guaranteed that the data is not contradictory, it is possible to build a configuration described in the statement.

Output

For each test case print one integer --- the minimum number of ropes that need to be cut to make the candy fall to the ground.

Example

input

Copy

复制代码

4

3

4 3

3 1

1 2

4

9 2

5 2

7 7

3 4

5

11 7

5 10

12 9

3 2

1 5

3

5 6

4 5

7 7

output

Copy

复制代码
2
2
3
0

解题说明:水题,此题分析一下就能发现,只要绳子长度小于高度,那就必须要切断才能让糖果落地,直接判断即可。

cpp 复制代码
#include<stdio.h>
int main()
{
	int m, n, a, b, ans;
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
	{
		ans = 0;
		scanf("%d", &m);
		for (int j = 0; j < m; j++)
		{
			scanf("%d %d", &a, &b);
			if (a > b)
			{
				ans++;
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}
相关推荐
liulilittle11 分钟前
深度剖析:OPENPPP2 libtcpip 实现原理与架构设计
开发语言·网络·c++·tcp/ip·智能路由器·tcp·通信
88号技师18 分钟前
2025年6月一区-田忌赛马优化算法Tianji’s horse racing optimization-附Matlab免费代码
开发语言·算法·matlab·优化算法
ゞ 正在缓冲99%…1 小时前
leetcode918.环形子数组的最大和
数据结构·算法·leetcode·动态规划
十年编程老舅1 小时前
跨越十年的C++演进:C++20新特性全解析
c++·c++11·c++20·c++14·c++23·c++17·c++新特性
Kaltistss2 小时前
98.验证二叉搜索树
算法·leetcode·职场和发展
知己如祭2 小时前
图论基础(DFS、BFS、拓扑排序)
算法
mit6.8242 小时前
[Cyclone] 哈希算法 | SIMD优化哈希计算 | 大数运算 (Int类)
算法·哈希算法
c++bug2 小时前
动态规划VS记忆化搜索(2)
算法·动态规划
哪 吒2 小时前
2025B卷 - 华为OD机试七日集训第5期 - 按算法分类,由易到难,循序渐进,玩转OD(Python/JS/C/C++)
python·算法·华为od·华为od机试·2025b卷