Discription
Gildong owns a bulgogi restaurant. The restaurant has a lot of customers, so many of them like to make a reservation before visiting it.
Gildong tries so hard to satisfy the customers that he even memorized all customers’ preferred temperature ranges! Looking through the reservation list, he wants to satisfy all customers by controlling the temperature of the restaurant.
The restaurant has an air conditioner that has 3 states: off, heating, and cooling. When it’s off, the restaurant’s temperature remains the same. When it’s heating, the temperature increases by 1 in one minute. Lastly, when it’s cooling, the temperature decreases by 1 in one minute. Gildong can change the state as many times as he wants, at any integer minutes. The air conditioner is off initially.
Each customer is characterized by three values: ti — the time (in minutes) when the i-th customer visits the restaurant, li — the lower bound of their preferred temperature range, and hi — the upper bound of their preferred temperature range.
A customer is satisfied if the temperature is within the preferred range at the instant they visit the restaurant. Formally, the i-th customer is satisfied if and only if the temperature is between li and hi (inclusive) in the ti-th minute.
Given the initial temperature, the list of reserved customers’ visit times and their preferred temperature ranges, you’re going to help him find if it’s possible to satisfy all customers.
Input
Each test contains one or more test cases. The first line contains the number of test cases q (1≤q≤500). Description of the test cases follows.
The first line of each test case contains two integers n and m (1≤n≤100, −109≤m≤109), where n is the number of reserved customers and m is the initial temperature of the restaurant.
Next, n lines follow. The i-th line of them contains three integers ti, li, and hi (1≤ti≤109, −109≤li≤hi≤109), where ti is the time when the i-th customer visits, li is the lower bound of their preferred temperature range, and hi is the upper bound of their preferred temperature range. The preferred temperature ranges are inclusive.
The customers are given in non-decreasing order of their visit time, and the current time is 0.
Output
For each test case, print “YES” if it is possible to satisfy all customers. Otherwise, print “NO”.
You can print each letter in any case (upper or lower).
Example
input
4
3 0
5 1 2
7 3 5
10 -1 0
2 12
5 7 10
10 16 20
3 -100
100 0 0
100 -50 50
200 100 100
1 100
99 -100 0
output
YES
NO
YES
NO
Note
In the first case, Gildong can control the air conditioner to satisfy all customers in the following way:
At 0-th minute, change the state to heating (the temperature is 0).
At 2-nd minute, change the state to off (the temperature is 2).
At 5-th minute, change the state to heating (the temperature is 2, the 1-st customer is satisfied).
At 6-th minute, change the state to off (the temperature is 3).
At 7-th minute, change the state to cooling (the temperature is 3, the 2-nd customer is satisfied).
At 10-th minute, the temperature will be 0, which satisfies the last customer.
In the third case, Gildong can change the state to heating at 0-th minute and leave it be. Then all customers will be satisfied. Note that the 1-st customer’s visit time equals the 2-nd customer’s visit time.
In the second and the fourth case, Gildong has to make at least one customer unsatisfied.
题意
输入t，表示有t组样例。
每组样例输入n和m。n表示人数，m表示初始温度。
下面n行，每行输入三个数，t，l，h。
ti表示进入时间，li表示第i位的最低温度，hi表示最高温度。
已知每分钟只能升高或者降低1度。
思路
根据时间差计算每个温度在某时刻的的温度范围，看要求的温度时候与实际范围有重叠，如果没有就是不可以，然后将温度范围缩小到最小的温度范围。
AC代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define pd(n) printf("%d\n", (n))
#define pdd(n, m) printf("%d %d\n", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define rep(i, a, n) for(int i=a; i<=n; i++)
#define per(i, n, a) for(int i=n; i>=a; i--)
int T;
int f;
int n;
ll m;
ll t[110],l[110],h[110];
int main()
{
scanf("%d",&T);
while(T--)
{
f=0;
sd(n);
sld(m);
ll low=m,high=m;
for(int i=1;i<=n;i++)
slddd(t[i],l[i],h[i]);
t[0]=0;
for(int i=1;i<=n;i++)
{
low-=(t[i]-t[i-1]);
high+=(t[i]-t[i-1]);
if(low<l[i])
low=l[i];
if(high>h[i])
high=h[i];
if(high-low<0)
{
f=1;
break;
}
}
if(f)
cout<<"NO"<<endl;
else
cout<<"YES"<<endl;
}
return 0;
}
