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 = 0;
    vector<int> dst2src, re;
    scanf("%d", &N);
    int temp = 0;
    dst2src.push_back(temp);
    for(int i = 0; i < N; i++){
        scanf("%d", &temp);
        sum += temp;
        dst2src.push_back(sum);
    }
    scanf("%d", &M);
    for(int i = 0; i < M; i++){
        int a, b, L1, L2;
        scanf("%d%d", &a, &b);
        if(a > b) swap(a, b);
        L1 = dst2src[b - 1] - dst2src[a - 1];
        L2 = sum - L1;
        printf("%d
", min(L1, L2));
    }
    return 0;
}

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。