1046. Shortest Distance

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 … 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.

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

import java.util.Scanner;public class Main {    public static void main(String[] args) {        Scanner sc = new Scanner(System.in);        int n = sc.nextInt();        int[] dist = new int[1+n];        int sum = 0;        for(int i=2; i<1+n; i++) {            sum += sc.nextInt();            dist[i] = sum;        }        sum += sc.nextInt();        int half = sum / 2;        int m = sc.nextInt();               for(int i=0; i<m; i++) {            int s = sc.nextInt();            int t = sc.nextInt();            int d1 = 0;            if(s > t)                d1 = dist[s] - dist[t];            else                d1 = dist[t] - dist[s];            if(d1 <= half)                System.out.println(d1);            else                System.out.println(sum - d1);        }    }}

#include <iostream>  #include <vector>  #include <stdio.h>  #include <math.h>  using namespace std;  vector<int> dis;  vector<int> disToOrigin;  int sum = 0;  int computeMinDis(int s, int e){      int minDist1 = 0,minDist2 = 0;      int start, end;      if(s < e) { start = s; end = e; }      else if( s == e ) return 0;      else { start = e; end = s; }      minDist1 = disToOrigin[end] - disToOrigin[start];      minDist2 = sum - minDist1;      return minDist1 < minDist2 ? minDist1 : minDist2;  }  int main()  {      int N,M;      cin >> N;      dis.resize(N+1);      disToOrigin.resize(N+1);      disToOrigin[1] = 0;      int value;      for(int i = 1; i <= N; i++){          scanf("%d",&value);          dis[i] = value;          if(i > 1) disToOrigin[i] = sum;          sum += value;      }      cin >> M;      int x,y;      for(int i = 0; i < M; i++){          scanf("%d%d",&x,&y);          printf("%d\n",computeMinDis(x,y));      }      return 0;  } 

我表示内心是崩溃的，Java一直运行超时，然而和C++算法都是一样的，

由此可见Java和C++的运行效率的差别

刷PAT真是好蛋疼啊，别人全部刷完时有多屌？？？？