1046 Shortest Distance （20 分）

最新推荐文章于 2020-03-26 21:18:07 发布

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本文介绍了一个简单的算法，用于计算高速公路形成简单环形时任意两个出口之间的最短距离。输入包括出口数量、各出口间的距离及待求距离的出口对，输出为每对出口间的最短距离。

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 D1D2 ⋯ 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

C：

首尾距离记录法：

/*
 @Date    : 2017-12-08 09:18:55
 @Author  : 酸饺子 (changzheng300@foxmail.com)
 @Link    : https://github.com/SourDumplings
 @Version : $Id$
*/

/*
https://www.patest.cn/contests/pat-a-practise/1046
 */

#include <stdio.h>
#include <stdlib.h>

#define MAXN 100000
#define MAXM 10000
#define min(x, y) (x) < (y) ? (x) : (y)

int main()
{
    int N, i;
    int D[MAXN], sdist[MAXN], edist[MAXN];
    scanf("%d", &N);
    for (i = 0; i < N; i++)
    {
        scanf("%d", &D[i]);
        if (i == 0)
        {
            sdist[i] = 0;
        }
        else
        {
            sdist[i] = sdist[i-1] + D[i-1];
        }
    }
    for (i = N - 1; i >= 0; i--)
    {
        if (i == N - 1)
        {
            edist[i] = 0;
        }
        else
        {
            edist[i] = edist[i+1] + D[i];
        }
    }

    int f, b, s, e;
    int M;
    int result[MAXM];
    scanf("%d", &M);
    for (i = 0; i < M; i++)
    {
        scanf("%d %d", &s, &e);
        s -= 1;
        e -= 1;
        if (s <= e)
        {
            f = sdist[e] - sdist[s];
            b = edist[e] + D[N-1] + sdist[s];
        }
        else
        {
            f = edist[s] + D[N-1] + sdist[e];
            b = sdist[s] - sdist[e];
        }
        result[i] = min(f, b);
    }

    for (i = 0; i < M; i++)
    {
        printf("%d\n", result[i]);
    }

    return 0;
}