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

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本文介绍了一道关于计算环形高速公路上任意两个出口间最短距离的问题。通过给出的相邻出口间的距离，利用前缀和的方法快速计算出所有询问的最短路径。

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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 ... D_N, where D_i is the distance between the i-th and the (i+1)-st exits, and D_N 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;
}