PAT 1046 Shortest Distance

最新推荐文章于 2025-03-14 11:15:13 发布

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最新推荐文章于 2025-03-14 11:15:13 发布

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本文介绍了一种算法，用于解决给定N个形成简单环形高速公路出口时，任意两个出口间的最短距离问题。输入包括出口数量、各段距离及询问对，输出为每对出口间的最短距离。

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The task is really simple: given N exits on a highway which forms a simple cycle, you are supposed to tell the shortest distance between any pair of exits.

Input Specification:

Each input file contains one test case. For each case, the first line contains an integer N (in [3,105]), followed by N integer distances D1 D2 ⋯ DN, where Di is the distance between the i-th and the (i+1)-st exits, and DN is between the N-th and the 1st exits. All the numbers in a line are separated by a space. The second line gives a positive integer M (≤104), with M lines follow, each contains a pair of exit numbers, provided that the exits are numbered from 1 to N. It is guaranteed that the total round trip distance is no more than 107.

Output Specification:

For each test case, print your results in M lines, each contains the shortest distance between the corresponding given pair of exits.

Sample Input:

5 1 2 4 14 9
3
1 3
2 5
4 1

Sample Output:

3
10
7

题意:N个点组成一个环，问任意两点之间最近的距离是多少.Di 表示从i位置到i+1的距离

思路:1.任意两点之间有两种走法，分别是顺时针和逆时针走.

2.在计算距离之前，我们不妨先处理从点1到任意点n之间的距离(n>1)，可以得出1-n的距离(数组的前缀和)

dis[n]=dis[n-1]+D[n-1]

但既然是一个环，那么从点N到点1的距离也要求出，修改dis[1]的值 :dis[1]=dis[N]+D[N]

3.当询问点x,y时，存在两种情况

A.x<=y

B.x>y

分别求出顺时针方向和逆时针方向的距离取较小值即可

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int Max=1e5+5;
int main()
{
    int N,Arr[Max],Pri[Max];
    scanf("%d",&N);
    Arr[0]=Pri[0]=0;
    for(int i=1;i<=N;i++)
    {
        scanf("%d",&Arr[i]);
        Pri[i]=Arr[i-1]+Pri[i-1];
    }
    Pri[1]=Arr[N]+Pri[N];
    int M,u,v;
    scanf("%d",&M);
    while(M--)
    {
        scanf("%d %d",&u,&v);
        if(u<=v)
        {
            if(u==1)
                printf("%d\n",min(Pri[v],Pri[1]-Pri[v]));
            else
                printf("%d\n",min(Pri[v]-Pri[u],Pri[1]-Pri[v]+Pri[u]));
        }
        else
        {
            if(v==1)
                printf("%d\n",min(Pri[u],Pri[1]-Pri[u]));
            else
                printf("%d\n",min(Pri[u]-Pri[v],Pri[1]-Pri[u]+Pri[v]));
        }
    }
    return 0;
}