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NC24445. [USACO 2014 Dec B]Marathon

描述

Unhappy with the poor health of his cows, Farmer John enrolls them in an assortment of different physical fitness activities. His prize cow Bessie is enrolled in a running class, where she is eventually expected to run a marathon through the downtown area of the city near Farmer John's farm! The marathon course consists of N checkpoints (3 <= N <= 100,000) to be visited in sequence, where checkpoint 1 is the starting location and checkpoint N is the finish. Bessie is supposed to visit all of these checkpoints one by one, but being the lazy cow she is, she decides that she will skip up to one checkpoint in order to shorten her total journey. She cannot skip checkpoints 1 or N, however, since that would be too noticeable. Please help Bessie find the minimum distance that she has to run if she can skip up to one checkpoint. Note that since the course is set in a downtown area with a grid of streets, the distance between two checkpoints at locations (x1, y1) and (x2, y2) is given by |x1-x2| + |y1-y2|. This way of measuring distance -- by the difference in x plus the difference in y -- is sometimes known as "Manhattan" distance because it reflects the fact that in a downtown grid, you can travel parallel to the x or y axes, but you cannot travel along a direct line "as the crow flies".

输入描述

The first line gives the value of N.
The next N lines each contain two space-separated integers, x and y,
representing a checkpoint (-1000 <= x <= 1000, -1000 <= y <= 1000).
The checkpoints are given in the order that they must be visited.
Note that the course might cross over itself several times, with
several checkpoints occurring at the same physical location.  When
Bessie skips such a checkpoint, she only skips one instance of the
checkpoint -- she does not skip every checkpoint occurring at the same
location.

输出描述

Output the minimum distance that Bessie can run by skipping up to one
checkpoint.  Don't forget to end your output with a newline.  In the
sample case shown here, skipping the checkpoint at (8, 3) leads to the
minimum total distance of 14.

示例1

输入: 

4
0 0
8 3
11 -1
10 0

输出: 

14

原站题解

上次编辑到这里，代码来自缓存 点击恢复默认模板

C++14(g++5.4) 解法, 执行用时: 44ms, 内存消耗: 1156K, 提交时间: 2020-08-23 11:14:38 

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
int n,x[100005],y[100005];
int judge(int i,int j){
    return abs(x[i]-x[j])+abs(y[i]-y[j]);
}
int main(){
    cin>>n;
    int sum=0;
    for(int i=1;i<=n;i++){
        cin>>x[i]>>y[i];
    }
    for(int i=2;i<=n;i++){
        sum+=judge(i-1,i);
    }
    int maxx=-1;
    for(int i=2;i<n;i++){
        maxx=max(maxx,judge(i-1,i)+judge(i,i+1)-judge(i-1,i+1));
    }
    cout<<sum-maxx<<endl;
    return 0;
}

Python3(3.5.2) 解法, 执行用时: 722ms, 内存消耗: 28720K, 提交时间: 2020-04-26 14:07:51 

N=int(input())
arr=[list(map(int,input().split())) for i in range(N)]
skip_dis=[0 for i in range(N)]
sumdis=0
max_skip=0
for i in range(1,N) :
    sumdis+=abs(arr[i][0]-arr[i-1][0])+abs(arr[i][1]-arr[i-1][1])
    if i<=N-2 :
        tmp1=abs(arr[i][0]-arr[i-1][0])+abs(arr[i][1]-arr[i-1][1])+abs(arr[i][0]-arr[i+1][0])+abs(arr[i][1]-arr[i+1][1])
        tmp2=abs(arr[i+1][0]-arr[i-1][0])+abs(arr[i+1][1]-arr[i-1][1])
        max_skip=max(tmp1-tmp2,max_skip)

print(sumdis-max_skip)

C++11(clang++ 3.9) 解法, 执行用时: 74ms, 内存消耗: 1368K, 提交时间: 2020-04-28 22:28:42 

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
int n,x[100005],y[100005];
int judge(int i,int j){
	return abs(x[i]-x[j])+abs(y[i]-y[j]);
}
int main(){
	cin>>n;
	int sum=0;
	for(int i=1;i<=n;i++){
		cin>>x[i]>>y[i];
	}
	for(int i=2;i<=n;i++){
		sum+=judge(i-1,i);
	}
	int maxx=-1;
	for(int i=2;i<n;i++){
		maxx=max(maxx,judge(i-1,i)+judge(i,i+1)-judge(i-1,i+1));
	}
	cout<<sum-maxx<<endl;
	return 0;
}

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