PAT A1046 Shortest Distance (20)

题目地址：https://www.patest.cn/contests/pat-a-practise/1046

题目描述：

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.

输入格式（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.

输出格式（Output Specification）:

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

输入样例（Sample Input）:

5 1 2 4 14 9

3

1 3

2 5

4 1

输出样例（Sample Output）:

3

10

7

题意：

有Ｎ个结点围城一个圈，相邻两个点之间的距离已知，且每次只能移动到相邻点。然后给出Ｍ个询问，每个询问给出两个数字Ａ和Ｂ即结点编号（１＜＝Ａ，Ｂ＜＝Ｎ），求从Ａ号结点到Ｂ号结点的最短距离。

样例解释：

如图所示，共有５个结点，分别标号为１、２、３、４，相邻两点的距离在图上给出。总共三个询问：

１３：从１号点到３号点的最短距离为３，路径为１－＞２－＞３；

２５：从２号点到５号点的最短距离为１０，路径为２－＞１－＞５；

４１：从４号点到１号点的最短距离为７，路径为４－＞３－＞２－＞１。

解题思路：

步骤 1：以dis[ i ]表示 1 号结点按顺时针方向到达”i 号结点顺时针方向的下一个结点“的距离（1 <= i <= N），sum 表示一圈的总距离。于是对每个查询 left ->right，其结果就是 dis( left , right ) 与 sum - dis（left , right）中的较小值。

步骤 2：dis 数组和 sum 在读入时就可以进行累加得到。这样对每个查询 left -> right，dis（left，right）其实就是 dis[ right - 1] - dis[left - 1]。这样可以做到查询复杂度为O（1）。

注意：

查询的两个点的编号可能会有 left > right 的情况。这样情况下，需要交换 left 和 right

此题若没有经过预处理 dis 数组和 sum 的做法会很容易超时

之所以不把 dis[ i ] 设置为 1 号结点按顺时针方向到达 i 号结点的距离，是因为 N 号结点到达 1 号结点的距离无法被这个数组所保存

C++完整代码如下：

#include<cstdio>
#include<algorithm>
using namespace std;

const int MAXN = 100005;
int dis[MAXN], A[MAXN];    // dis数组含义已说明，A[I] 存放 i 号与 i+1 号顶点的距离
int main(){
int sum = 0, query, n, left, right;
scanf("%d", &n);
for(int i = 1; i <= n; i++){
scanf("%d", &A[i]);
sum += A[i];    //累加 sum
dis[i] = sum;     //预处理 dis 数组
}
scanf("%d", &query);
for(int i = 0; i < query; i++){      //query 个查询
scanf("%d%d", &left, &right);    // left -> right
if(left > right) swap(left, right);  //left -> right 时交换
int temp = dis[right - 1] - dis[left - 1];
printf("%d\n", min(temp, sum - temp));
}
return 0;
}