A1046. 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 ... D_N, where D_i is the distance between the i-th and the (i+1)-st exits, and D_Nis 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

 #include<cstdio>
 #include<iostream>
 #include<vector>
 #include<algorithm>
 using namespace std;
 int main(){
     int N, M, sum = ;
     vector<int> dst2src, re;
     scanf("%d", &N);
     int temp = ;
     dst2src.push_back(temp);
     for(int i = ; i < N; i++){
         scanf("%d", &temp);
         sum += temp;
         dst2src.push_back(sum);
     }
     scanf("%d", &M);
     for(int i = ; i < M; i++){
         int a, b, L1, L2;
         scanf("%d%d", &a, &b);
         if(a > b) swap(a, b);
         L1 = dst2src[b - ] - dst2src[a - ];
         L2 = sum - L1;
         printf("%d\n", min(L1, L2));
     }
     return ;
 }

总结：1、可使用swap、min函数，引用algorithm即可。

　　　2、预处理：模拟题一般要考虑到时间复杂度，本题中有N个节点。若不做预处理，M组查询，会有M*N的复杂度。若在读入数据的时候，就计算出源点到该点的距离，则在查询时，a到b的距离可用a到源减去b到源，不用遍历。

　　　3、由于是一个环，a到b的距离只有顺逆时针两种可能。