HDU - 2586 - How far away ? （最短路）

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本文详细解析HDU-2586-Howfaraway?问题，探讨在一个带权无向图中求解任意两点间最短路径的有效算法。通过对比Floyd算法与Dijkstra算法，最终采用深度优先搜索（DFS）实现简单粗暴的解决方案，适用于数据量适中的场景。

HDU - 2586 - How far away ? （最短路）

There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this “How far is it if I want to go from house A to house B”? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path(“simple” means you can’t visit a place twice) between every two houses. Yout task is to answer all these curious people.
Input
First line is a single integer T(T<=10), indicating the number of test cases.
For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.
Output
For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.
Sample Input
2
3 2
1 2 10
3 1 15
1 2
2 3

2 2
1 2 100
1 2
2 1
Sample Output
10
25
100
100

题目大意：

给你一个带权无向图，有n个顶点，n-1条边。然后有m次询问，每次询问有 a b两个数，需要输出a b 之间的最短路。

抽象一下，就是求任意两点之间的最短路。

解题思路：

任意两点最短路？？？首先想到的肯定是Floyd算法，但是。。。数据量n=1e4…复杂度0（n³）肯定是不行的
然后我就考虑把询问里的数按出现次序排序，然后先用Dijkstra跑出现次数最多的顶点w为起点的最短路，然后把结果用个标记数组 book[i][j]记录下来，然后遇到已经跑过的最短路，也就是book数组里有的就直接输出，然后每次都跑Dijkstra。。。虽然i想的很美好，但是却给我显示超内存了？？？

无奈之下，上网搜了一波题解
题解上用dfs每次询问，找一下最短路。。。简单粗暴又实用。。。数据量是1e4 边数也是1e4的数据量询问数<100…所以也不会超时。。

搜索的时候我们用个标记数组vis[]表示这个点是否经过过，然后遇到中终点就输出结果，不是终点就继续dfs。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=40009;
int vis[maxn];
int head[maxn];
int cnt=0;
int n,m;
struct node{
	int to;
	int val;
	int next;
}side[maxn*2];
void init()
{
	memset(head,-1,sizeof(head));
	memset(vis,0,sizeof(vis));
	cnt=0;
}
void add(int x,int y,int w)
{
	side[cnt].to=y;
	side[cnt].val=w;
	side[cnt].next=head[x];
	head[x]=cnt++;
}
void dfs(int a,int b,int step)
{
	vis[a]=1;
	for(int i=head[a];i!=-1;i=side[i].next)
	{
		int k=side[i].to;
		if(vis[k])
			continue;
		if(k==b)
		{
			cout<<step+side[i].val<<endl;
			return ;
		 } 
		 else
		 {
		 	dfs(k,b,step+side[i].val);
		 }
	}
}
int main()
{
	int T;
	cin>>T;
	while(T--)
	{
		cin>>n>>m;
		init();
		for(int i=0;i<n-1;i++)
		{
			int a,b,c;
			scanf("%d %d %d",&a,&b,&c);
			add(a,b,c);
			add(b,a,c);
		}
		while(m--)
		{
			int a,b;
			cin>>a>>b;
			memset(vis,0,sizeof(vis));
			dfs(a,b,0);
		}
	}
	return 0;
}