1046. Shortest Distance (20)

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 14 931 32 54 1

Sample Output:

3107

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给出一个环和其中相邻点之间的距离，求指定的两点之间的最短距离。两点之间，走法只有两种，即顺时针和逆时针走两种走法，选使得路径最短的那种就行了。输入的时候计算环的总长度sum，同时保存点1到各个点的距离dis[i]（i表示当前点）。然后输入需要查找的两个点x和y，使x小于y（如果x大于y则互换），用dis[y]减去dis[x]即可得到x到y的顺时针距离，然后sum减去顺时针距离就得到逆时针距离，然后他们之间的最小值就是答案。

代码：

#include <iostream>#include <cstring>#include <vector>#include <cstdio>#include <cstdlib>using namespace std;int main(){int n;cin>>n;vector<int>dis(n+1);int sum=0;for(int i=1;i<=n;i++){int d;cin>>d;sum+=d;if(i==n) break;dis[i+1]=dis[i]+d;}int m;cin>>m;for(int i=0;i<m;i++){int x,y;cin>>x>>y;if(x>y) swap(x,y);int res=min((dis[y]-dis[x]),sum-(dis[y]-dis[x]));cout<<res<<endl;}}