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 [3,10^5]), 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 (≤10^4), 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 10^7.
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
题目大意
一个甜甜圈，求该圈上标准的任意两点间，抵达的最短距离
思路
记甜甜圈的总长为sum，length[z]表示点z抵达点1的距离，lebgth[1] = 0
那么任意两点A与B间的最短距离为
min(sum-abs(length[A]-length[B]),abs(length[A]-length[B]))
C/C++
#include
using namespace std;
int main()
{
int N,length[100001]={0},sumLen=0,len;
cin >> N;
for(int z=1;z> len;
length[z+1] = sumLen = sumLen + len;
}
cin >> N;
while (N--){
int a,b;
cin >> a >> b;
int minRoad = abs(length[a]-length[b]);
minRoad = min(minRoad,sumLen-minRoad);
cout << minRoad << endl;
}
return 0;
}
Original: https://blog.csdn.net/daybreak_alonely/article/details/127824714
Author: 三块不一样的石头
Title: 1046 Shortest Distance
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