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

题意：给你一个环形路上相邻两点的距离，求任意两点之间的最短路

解题思路：求出距离的前缀和即可

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <map>

using namespace std;

#define ll long long
int n,k;
int sum[100090];

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