PAT 甲级 1046 Shortest Distance

最新推荐文章于 2024-06-02 14:13:57 发布

drizzly丶

最新推荐文章于 2024-06-02 14:13:57 发布

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本文链接：https://blog.youkuaiyun.com/x_zjudream/article/details/85645990

本文介绍了一个简单的算法，用于计算高速公路形成简单环形上的任意两个出口之间的最短距离。通过输入环形高速上各出口间的距离，算法能快速计算并输出任意一对出口间的最短路径长度。

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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

刚开始做有个测试点老是超时。。。

后来加了个数组记录边的累积和

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int dis[maxn],an[maxn];
int main()
{
int max=0;
int n,a,b,k;
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>dis[i];
max+=dis[i];
an[i] = max;
}
cin>>k;
while(k--)
{
int sum=0;
cin>>a>>b;
if(b<a)
swap(a,b);
sum = an[b-1]-an[a-1];
cout<<min(sum,max-sum)<<endl;
}
}