最短路径（shortest distance）

#include<cstdio>#include<iostream>#include<cstring>#include<cmath>#include<cstdlib>using namespace std;struct node{int num;int length;};int main(){node* a;a=(node*)malloc(100000*sizeof(node));int n=0;cin>>n;int sum=0;for(int i=0;i<n;i++){int m=0;cin>>m;(a+i)->num=i+1;(a+i)->length=m;sum+=m;}//sum+=(a+i+1)->length;int k=0;cin>>k;for(int j=0;j<k;j++){int k1=0,k2=0;cin>>k1>>k2;int _sum=0;if(k1>k2){int temp=k2;k2=k1;k1=temp;}for(int t=k1-1;t<k2-1;t++){_sum+=(a+t)->length;}if(_sum>sum-_sum){cout<<sum-_sum<<endl;}else{cout<<_sum<<endl;}}cin.get();return 0;}

1046. Shortest Distance (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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