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

求两点间最短距离。在输入的时候把前N项的和存储起来，最后直接相减

#include<iostream>  #include<cstring>  #include<cstdio>  #include<queue>  #include<stack>  #include<algorithm>  #include<vector> #include<map>using namespace std;int main(){int n;cin>>n;int sum[100001]={0};for(int i=1;i<=n;i++){int num;scanf("%d",&num);sum[i+1]=sum[i]+num;}int m;cin>>m;int out=0,mm=sum[n+1];for(int i=0;i<m;i++){int s,e;scanf("%d %d",&s,&e);int mi=s<e?s:e;int ma=s>e?s:e;if(sum[ma]-sum[mi]<mm-(sum[ma]-sum[mi])){out=sum[ma]-sum[mi];}else out=mm-(sum[ma]-sum[mi]);cout<<out<<endl;}return 0;}