一、题目描述
原题链接
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:
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
二、解题思路
题目大意是：N个出口构成一个圈，给出每个出口离下个出口的距离，随后任意输入两个出口，要我们输出它们之间的最短距离。这道题目很好想，因为每个出口构成一个圈，我们的最短距离要么是按顺时针方向转过去，要么是按逆时针方向转过去。我们可以用数组fromSt表示每个出口离起点的距离（顺时针），那么，根据题目给出的两个出口，我们可以求出顺时针的距离，随后用整个圆的周长减去这个距离就是逆时针方向的距离，进行比较然后输出即可。
三、AC代码
#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn = 100010;
int edge[maxn] = {
0}; //存放第i条边的长度
int fromSt[maxn] = {
0};
int N;
int main()
{
int M, a, b, ans;
scanf("%d", &N);
for(int i=1; i<=N; i++)
{
scanf("%d", &edge[i]);
fromSt[i] = fromSt[i-1] + edge[i-1];
}
scanf("%d", &M);
for(int i=0; i<M; i++)
{
scanf("%d%d", &a, &b);
int clock = fromSt[max(a, b)] - fromSt[min(a, b)];
int counter = fromSt[N] + edge[N] - clock;
ans = clock > counter ? counter : clock;
printf("%d\n", ans);
}
return 0;
}