(Jizhong) 2189. Marathon

(File IO): input: marathon.in output: marathon.out
time limit: 1000 ms space constraints: 262144 KB specific restrictions
Goto ProblemSet

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Title Description
There on the map $N$ cities, a cow from $1$ Number City sequentially through $N$ cities, and finally to $N$ number of cities. But only cows think this is too boring, so it decided to skip a city which (but not skip $1$ number and $N$ number change), such that it from $1$ Number City began arriving $N$ minimum number of cities through which the total distance. Every city has a coordinate, from the city $(x1, y1)$ to the city $(x2, y2)$ a distance $|x1 - x2| + |y1 - y2|$。

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Enter
a number of the first row $N$ , is the number of cities
on the next line $N$ each row two rows $x, y$ , represent the coordinates of each city

Output
line a number $years$ , such that it from $1$ Number City began to skip a certain city, arrival $N$ number of cities through which the smallest total distance

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Sample input
. 4
0 0
. 8. 3
. 11 -1
10 0

Sample output
14

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Data range limit
• For $40$ % of the data, $N <= 1000$ .
• For $100$ % of the data, $3 <= N <= 10^5，-10^3 <= x <= 10^3，-10^3 <= y <= 10^3$。

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Tips
skip $2$ Hao City

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Problem-solving ideas
direct violence enumeration. . .

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Code

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<iomanip>
#include<cmath>
using namespace std;
int n,a[100010][3],n,minn,ans,t;
int main(){
	freopen("marathon.in","r",stdin);
    freopen("marathon.out","w",stdout);
    scanf("%d",&n);
    minn=2147483647;
    for(int i=1;i<=n;i++)
    {
    	scanf("%d%d",&a[i][1],a[i][2]);
    	t=t+abs(a[i-1][1]-a[i][1])+abs(a[i-1][2]-a[i][2]);
	}
	for(int i=1;i<=n;i++)
	{
		ans=s-(abs(a[i][1]-a[i-1][1])+abs(a[i][2]-a[i-1][2]))-(abs(a[i][1]-a[i+1][1])+abs(a[i][2]-a[i+1][2]))+(abs(a[i-1][1]-a[i+1][1])+abs(a[i-1][2]-a[i+1][2]));
    	if(ans<minn)
    		minn=ans;
	}
	printf("%d",minn);
}