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
佳児素花痴╮1 小时前
树的基础知识与查找排序算法
数据结构·算法
郝学胜-神的一滴2 小时前
Python 高级编程 026:内置数据结构之骈文纵论
开发语言·数据结构·python·程序人生·软件工程
程序员老舅8 小时前
啃透 I2C 驱动开发,才算入门嵌入式 Linux 内核驱动
数据结构·驱动开发·b树·内核·嵌入式·嵌入式开发·i2c
lueluelue478 小时前
LeetCode:链表
算法·leetcode·链表
橘子汽水16811 小时前
Leetcode 23,543合并K个升序链表,二叉树的直径
算法·leetcode·链表
huameinan狮子11 小时前
Adaboost算法原理与计算实例
算法
Tri_Function11 小时前
动态规划DP1(c++)
c++
别怪我很水11 小时前
API中提供了VelocityTracker类用于计算触摸事件MotionEvent的速度,而其内部默认使用的方法就是最小二乘法,本 ...
算法·gitee·最小二乘法
清水迎朝阳14 小时前
客户端软件 — 用户统计方案
服务器·c++·客户端·用户统计