CodeForces 709B Checkpoints
2016-09-10 08:26
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B. Checkpoints
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Vasya takes part in the orienteering competition. There are n checkpoints located along the line at coordinates x1, x2, ..., xn.
Vasya starts at the point with coordinate a. His goal is to visit at least n - 1 checkpoint
in order to finish the competition. Participant are allowed to visit checkpoints in arbitrary order.
Vasya wants to pick such checkpoints and the order of visiting them that the total distance travelled is minimized. He asks you to calculate this minimum possible value.
Input
The first line of the input contains two integers n and a (1 ≤ n ≤ 100 000, - 1 000 000 ≤ a ≤ 1 000 000) —
the number of checkpoints and Vasya's starting position respectively.
The second line contains n integers x1, x2, ..., xn ( - 1 000 000 ≤ xi ≤ 1 000 000) —
coordinates of the checkpoints.
Output
Print one integer — the minimum distance Vasya has to travel in order to visit at least n - 1 checkpoint.
Examples
input
output
input
output
input
output
0
将n个点排序之后
n个点要走n-1个点,那么最后一个点肯定呢是边界两个点之一,分两种情况讨论
每种情况,可以根据起点和n-1个点两端的位置关系分类
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <algorithm>
#include <math.h>
using namespace std;
const int maxn=1e5;
int n,start;
int a[maxn+5];
int main()
{
scanf("%d%d",&n,&start);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
if(n==1)
{
printf("0\n");
return 0;
}
int res1,res2;
sort(a+1,a+n+1);
if(start<=a[2])
res1=abs(a
-start);
else if(start>=a
)
res1=abs(a[2]-start);
else
res1=min(abs(a[2]-start)+abs(a[2]-a
),abs(a
-start)+abs(a[2]-a
));
if(start<=a[1])
res2=abs(a[n-1]-start);
else if(start>=a[n-1])
res2=abs(a[1]-start);
else
res2=min(abs(a[1]-start)+abs(a[1]-a[n-1]),abs(a[n-1]-start)+abs(a[1]-a[n-1]));
printf("%d\n",min(res1,res2));
return 0;
}
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Vasya takes part in the orienteering competition. There are n checkpoints located along the line at coordinates x1, x2, ..., xn.
Vasya starts at the point with coordinate a. His goal is to visit at least n - 1 checkpoint
in order to finish the competition. Participant are allowed to visit checkpoints in arbitrary order.
Vasya wants to pick such checkpoints and the order of visiting them that the total distance travelled is minimized. He asks you to calculate this minimum possible value.
Input
The first line of the input contains two integers n and a (1 ≤ n ≤ 100 000, - 1 000 000 ≤ a ≤ 1 000 000) —
the number of checkpoints and Vasya's starting position respectively.
The second line contains n integers x1, x2, ..., xn ( - 1 000 000 ≤ xi ≤ 1 000 000) —
coordinates of the checkpoints.
Output
Print one integer — the minimum distance Vasya has to travel in order to visit at least n - 1 checkpoint.
Examples
input
3 10 1 7 12
output
7
input
2 0 11 -10
output
10
input
5 0 0 0 1000 0 0
output
0
将n个点排序之后
n个点要走n-1个点,那么最后一个点肯定呢是边界两个点之一,分两种情况讨论
每种情况,可以根据起点和n-1个点两端的位置关系分类
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <algorithm>
#include <math.h>
using namespace std;
const int maxn=1e5;
int n,start;
int a[maxn+5];
int main()
{
scanf("%d%d",&n,&start);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
if(n==1)
{
printf("0\n");
return 0;
}
int res1,res2;
sort(a+1,a+n+1);
if(start<=a[2])
res1=abs(a
-start);
else if(start>=a
)
res1=abs(a[2]-start);
else
res1=min(abs(a[2]-start)+abs(a[2]-a
),abs(a
-start)+abs(a[2]-a
));
if(start<=a[1])
res2=abs(a[n-1]-start);
else if(start>=a[n-1])
res2=abs(a[1]-start);
else
res2=min(abs(a[1]-start)+abs(a[1]-a[n-1]),abs(a[n-1]-start)+abs(a[1]-a[n-1]));
printf("%d\n",min(res1,res2));
return 0;
}
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