题目:
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 1Sample Output:
3 10 7
思路:
分为循环正向相加和反向相加两种选择,都进行计算,取最小值。
代码:
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
//ifstream cin;
//cin.open("case1.txt");
int dis[100001] = { 0 };
int total_dis[100001] = { 0 };
int N;
cin>>N;
int i;
for (i = 1; i <= N; ++i)
{
cin >> dis[i];
total_dis[i] = total_dis[i - 1] + dis[i];
}
int M,n1,n2,tmp,d1,d2;
cin >> M;
vector<vector<int>> p(M,vector<int>(2));
for (i = 0; i < M; ++i)
{
cin >> p[i][0] >> p[i][1];
}
for (i = 0; i < M; ++i)
{
n1 = p[i][0];
n2 = p[i][1];
if (n2 < n1)
{
tmp = n2;
n2 = n1;
n1 = tmp;
}
d1 = total_dis[n2 - 1] - total_dis[n1 - 1];
d2 = total_dis[N] - total_dis[n2 - 1];
d2 += total_dis[n1 - 1];
if (d1 < d2)
cout << d1;
else
cout << d2;
cout << endl;
}
system("pause");
return 0;
}