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, 10^5^]), followed by N integer distances D~1~ D~2~ ... 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 (<=10^4^), 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 10^7^.

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

题目大意：给出一串数字(坐标从1开始)， 这些数字围成一个圈，找出给出亮点的最小距离；
第一次试的时候，超时了，因为处理的不是很完善
解法：1.输入的时候，把所有的值加起来，计算的时候，只需要计算一个方向的距离就行， 另一个方向的距离用总的距离（即一个环的长度）来减就行；这种方法在最后一个测试点不能通过，因为每次计算都要重复的计算一些点的距离；
　　　2.对一进行一些小的改进，把数组中的值，保存为距离第一个点的距离，这样在计算的时候，把两点离第一个点的距离想减，就得到两点的距离。 为了方便计算，添加一个下标为0的点。
描述点不是很清晰，还是看代码吧。
关键在于，在输入的时候，就对数据进行处理，减少循环次数

#include<iostream>
using namespace std;
int main(){
  int n, m, i, sum=0, *arr;
  cin>>n;
  arr = new int[n];
  for(i=0; i<n; i++){
    cin>>arr[i];
    sum += arr[i];
  }
  cin>>m;
  for(i=0; i<m; i++){
    int a, b, j, l=0;
    cin>>a>>b;
    if(a>b) swap(a, b);
    for(j=a; j<b; j++) l += arr[j-1];
    l = l<(sum-l)?l:(sum-l);
    cout<<l<<endl;
  }
  return 0;
}

修改之后点代码

#include<iostream>
using namespace std;
int main(){
  int n, m, i, sum=0, *arr;
  cin>>n;
  arr = new int[n+1];
  arr[0]=0;
  for(i=1; i<=n; i++){
    cin>>arr[i];
    sum += arr[i];
    arr[i] += arr[i-1];
  }
  cin>>m;
  for(i=0; i<m; i++){
    int a, b, l=0;
    cin>>a>>b;
    if(a>b) swap(a, b);
    l = arr[b-1]- arr[a-1];
    l = l<(sum-l)?l:(sum-l);
    cout<<l<<endl;
  }
  return 0;
}