目錄
題目描述
New Building for SIS
time limit per test 1 second
memory limit per test 256 megabytes
input
standard input
output
standard output
You are looking at the floor plan of the Summer Informatics School's new building. You were tasked with SIS logistics, so you really care about travel time between different locations: it is important to know how long it would take to get from the lecture room to the canteen, or from the gym to the server room.
The building consists of n towers, h floors each, where the towers are labeled from 1 to n, the floors are labeled from 1 to h. There is a passage between any two adjacent towers (two towers i and i + 1 for all i: 1 ≤ i ≤ n - 1) on every floor x, where a ≤ x ≤ b. It takes exactly one minute to walk between any two adjacent floors of a tower, as well as between any two adjacent towers, provided that there is a passage on that floor. It is not permitted to leave the building.
The picture illustrates the first example.
You have given k pairs of locations (ta, fa), (tb, fb): floor fa of tower ta and floor fb of tower tb. For each pair you need to determine the minimum walking time between these locations.
Input
The first line of the input contains following integers:
- n: the number of towers in the building (1 ≤ n ≤ 108),
- h: the number of floors in each tower (1 ≤ h ≤ 108),
- a and b: the lowest and highest floor where it's possible to move between adjacent towers (1 ≤ a ≤ b ≤ h),
- k: total number of queries (1 ≤ k ≤ 104).
Next k lines contain description of the queries. Each description consists of four integers ta, fa, tb, fb (1 ≤ ta, tb ≤ n, 1 ≤ fa, fb ≤ h). This corresponds to a query to find the minimum travel time between fa-th floor of the ta-th tower and fb-th floor of the tb-th tower.
Output
For each query print a single integer: the minimum walking time between the locations in minutes.
Example
input
3 6 2 3 3 1 2 1 3 1 4 3 4 1 2 2 3
output
1 4 2
題意分析
題意：給你一系列查詢，問從樓層A到樓層B最少要多久時間，橫跨豎行一格都只要1min
但是要橫跨樓層的話，就必須在所給範圍內纔可跨越。
一開始少考慮了一個條件WA了一發，首先是樓層不同，判斷一下是否在所給的範圍內
在外面逼近限定就好，然後就是在同一樓層，直接抵達即可 （WA點）。
AC代碼
#include <iostream>
#include <cstdio>
#include <algorithm>
#define MAXN 10005
using namespace std;
int main()
{
long long n,h,a,b,k; // a 最低 b 最高
long long sum=0;
long long q,w,e,r;
int i;
while(cin>>n>>h>>a>>b>>k)
{
for(i=0;i<k;i++)
{
sum=0;
cin>>q>>w>>e>>r;
if(q==e)
{
sum += abs(w-r);
}
else
{
sum += abs(q-e);
if(w>b)
{
sum += abs(w-b);
w=b;
}
if(w<a)
{
sum += abs(w-a);
w=a;
}
if(r>b)
{
sum += abs(r-b);
r=b;
}
if(r<a)
{
sum += abs(r-a);
r=a;
}
sum += abs(w-r);
}
cout<<sum<<endl;
}
}
return 0;
}