1046. Shortest Distance (20)

最新推荐文章于 2021-10-05 21:31:24 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。输入包括环形高速上的出口数量及各段距离，以及需要查询距离的出口对；输出则是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 D₁ D₂ ... D_N, where D_i is the distance between the i-th and the (i+1)-st exits, and D_N 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

记录路径的方法不会超时！

参考代码：

#include <iostream>
#include <vector>
#include <stdio.h>
using namespace std;
vector<int> roads(100001);
int N,M,totalLength;
int main()
{
    scanf("%d",&N);
    totalLength = 0;
    roads[0] = 0;
    for(int i=0;i<N;i++){
        int temp;
        scanf("%d",&temp);
        totalLength += temp;
        roads[i+1] = roads[i]+temp;
    }
    scanf("%d",&M);
    for(int i=0;i<M;i++){
        int start,last;
        scanf("%d%d",&start,&last);
        if(start>last){
            swap(start,last);
        }
        int sum,other,endS;
        sum = roads[last-1]-roads[start-1];
        other = totalLength-sum;
        endS = sum<other?sum:other;
        printf("%d\n",endS);
    }
    return 0;
}