算法笔记习题：Shortest Distance

最新推荐文章于 2025-08-05 19:50:13 发布

Bamboo9999

最新推荐文章于 2025-08-05 19:50:13 发布

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文章标签：算法 c语言数据结构

本文链接：https://blog.youkuaiyun.com/Bamboo9999/article/details/120478277

该博客讨论了一个计算机算法问题，即给定一个由N个节点构成的环形高速公路，计算任意两个节点间的最短路径。每个节点间存在距离，并且总距离不超过10^7。输入包括节点数N和各个节点间的距离，以及查询的节点对数M。解决方案通过遍历节点和考虑两种路径（直接路径和绕环路径）来找出最短距离。代码示例使用了C++，通过比较两种路径的长度输出最短距离。

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题目

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.

输入

Each input file contains one test case. For each case, the first line contains an integer N (in [3, 10^5]), 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 (<=10^4), 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.

输出

For each test case, print your results in M lines, each contains the shortest distance between the corresponding given pair of exits.

案例

5 1 2 4 14 9
3
1 3
3
2 5
10
4 1
7

题目解析

N个结点围成一个圈，求任意两个点之间的最短距离。
N表示有N个点，然后输入D1，D2……DN,例如Di就表示第i个点到第i+1个点的距离，DN存放的是第N个点到第一个点的距离；
然后输入M[≤10^4]，表示接下来有M组数据，求每组数据的最短距离

代码

#include <stdio.h>

int main(){
	int n,m;
	int a,b;

	// 数组大小要写对
	int d[100000];
	
	// 两个结点有两条路径，r1表示普通路径（例如：1->2->3），r2表示特殊路径(例如：3->4->1)
	int r1 = 0,r2 = 0;
	
	while(scanf("%d",&n) != EOF){
		
		for(int i =0;i<n;i++){
			scanf("%d",&d[i]);
		}
		
		scanf("%d",&m);
		for(int i=0;i<m;i++){
			scanf("%d %d",&a,&b);
			
			//如果前面一个比较大，记得换个顺序
			if(a > b){
				int tmp = a;
				a = b;
				b = tmp;
			}
			
			for(int k=a-1;k<b-1;k++){
				r1 = r1 + d[k];
			}
			
			// 特殊路径要分成两段来求解
			for(int k=b-1;k<n;k++){
				r2 = r2 + d[k];
			}
			for(int k = 0;k<a-1;k++){
				r2 = r2 + d[k];
			}
			
			if(r1 > r2){
			printf("%d\n",r2);
			}else{
			printf("%d\n",r1);
		}
		
		// 重置数据
		r1 = 0,r2 = 0;
		
		}
	}
	return 0;
}