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 ... DN, where Di is the distance between the i-th and the (i+1)-st exits, and DNis 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
 1 #include<cstdio>
 2 #include<iostream>
 3 #include<vector>
 4 #include<algorithm>
 5 using namespace std;
 6 int main(){
 7     int N, M, sum = 0;
 8     vector<int> dst2src, re;
 9     scanf("%d", &N);
10     int temp = 0;
11     dst2src.push_back(temp);
12     for(int i = 0; i < N; i++){
13         scanf("%d", &temp);
14         sum += temp;
15         dst2src.push_back(sum);
16     }
17     scanf("%d", &M);
18     for(int i = 0; i < M; i++){
19         int a, b, L1, L2;
20         scanf("%d%d", &a, &b);
21         if(a > b) swap(a, b);
22         L1 = dst2src[b - 1] - dst2src[a - 1];
23         L2 = sum - L1;
24         printf("%d\n", min(L1, L2));
25     }
26     return 0;
27 }
View Code

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。

 

posted @ 2018-01-16 22:13  ZHUQW  阅读(105)  评论(0)  编辑  收藏  举报