PAT 1046 Shortest Distance [模拟] [超时]

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该博客介绍了PAT 1046题目的解决方案，题目要求在形成简单环形的高速公路上，找到任意两个出口之间的最短距离。博主首先遇到了因复杂度为O(N^2)导致的超时问题，然后优化了代码，通过存储每个点到固定点的距离，并计算两种可能的路径来求解最短距离，实现了AC（Accepted）的代码。

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 Diis 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

我的超时代码：

题目是“一根筋”的最短路径，两个地点之间只有两条路线可供选择。但遗憾的是复杂度O(N2)，造成最后一个测试超时。

#include<stdio.h>

int main()
{
	int num,d[100005],sum = 0;
	int x,a,b,i,j;
	scanf("%d",&num);
	for(i=1;i<=num;i++)
	{
		scanf("%d",&d[i]);
		sum += d[i];
	}
	scanf("%d",&x);
	int t;
	for(i=0;i<x;i++)
	{
		int min = 0;
		scanf("%d%d",&a,&b);
		if(a>b) //此处距离没有方向，4到1和1到4的结果相同，升序方便后面操作
		{
			t = a;
			a = b;
			b = t;
		} 
		for(j=a;j<b;j++)
		    min += d[j];
        //一条路线是a到b，另一条是总长度-a到b，取最小
        printf("%d\n",min<sum-min?min:sum-min);
	}	
    
}

我的AC代码：

将每个点到顶点①的长度dis[]存取下来，任意两点a、b之间的距离有两种情况：1.dis[b] - dis[a] 2.反方向的路线，即总长度减去情况1的长度。

#include<stdio.h>
#define maxN 100005

int main()
{
	int num,d[maxN],dis[maxN],sum = 0;
	int x,a,b,i,j;
	scanf("%d",&num);
	for(i=1;i<=num;i++)
	{
		scanf("%d",&d[i]);
		sum += d[i];
        //dis记录每个点到顶点①的距离，但不包括顶点①，所以i+1开始
		dis[i+1] = sum;
	}
	scanf("%d",&x);
	int t;
	for(i=1;i<=x;i++)
	{
		int min = 0;
		scanf("%d%d",&a,&b);
		if(a>b)
		{
			t = a;
			a = b;
			b = t;
		} 
		min = dis[b] - dis[a];
		//printf("%d %d\n",min,sum-min);
        printf("%d\n",min<sum-min?min:sum-min);
	}	 
}