PAT (Advanced Level) Practise 1046 Shortest Distance (20)

1046. Shortest Distance (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 1

Sample Output:

3
10
7

计算一下前缀和，然后比较一下是直接走过去还是绕到n

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 5;
int n, m, x, sum[maxn], y, k;

int main()
{
scanf("%d", &n);
for (int i = 2; i <= n; i++) scanf("%d", &x), sum[i] = sum[i - 1] + x;
scanf("%d%d", &k, &m);
while (m--)
{
scanf("%d%d", &x, &y);
if (x > y) swap(x, y);
printf("%d\n", min(sum[y] - sum[x], sum[x] + k + sum
- sum[y]));
}
return 0;
}