POJ-1986 Distance Queries

Posted 2020-09-29 ISGuXing

Distance Queries

Time Limit: 2000MS	Memory Limit: 30000K
Total Submissions: 13987	Accepted: 4924
Case Time Limit: 1000MS

Description

Farmer John\'s cows refused to run in his marathon since he chose a path much too long for their leisurely lifestyle. He therefore wants to find a path of a more reasonable length. The input to this problem consists of the same input as in "Navigation Nightmare",followed by a line containing a single integer K, followed by K "distance queries". Each distance query is a line of input containing two integers, giving the numbers of two farms between which FJ is interested in computing distance (measured in the length of the roads along the path between the two farms). Please answer FJ\'s distance queries as quickly as possible!

Input

* Lines 1..1+M: Same format as "Navigation Nightmare"

* Line 2+M: A single integer, K. 1 <= K <= 10,000

* Lines 3+M..2+M+K: Each line corresponds to a distance query and contains the indices of two farms.

Output

* Lines 1..K: For each distance query, output on a single line an integer giving the appropriate distance.

Sample Input

7 6
1 6 13 E
6 3 9 E
3 5 7 S
4 1 3 N
2 4 20 W
4 7 2 S
3
1 6
1 4
2 6

Sample Output

13
3
36

Hint

Farms 2 and 6 are 20+3+13=36 apart. 

这题目别想多，跟方向没关系，LCA。

关键在于怎么将距离整出来，我们可以有这么一个方程，

           ans[id]=pre[v].len+pre[u].len-2*pre[Find(v)].len;
用一个图来表示就是这样：


pre[v].len就是1到3的距离，
pre[u].len就是1到4的距离。
pre[Find(v)].len就是1到2的距离。
这样就很好解决了所有的问题了。

还有就是，重要的事情说三遍：并查集要压缩路径！并查集要压缩路径！并查集要压缩路径！
之前几次忘了压缩，直接TLE；一脸懵逼

附上AC代码：

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<string.h>
 4 #include<algorithm>
 5 #include<vector>
 6 using namespace std;
 7 struct Node{
 8     int f;
 9     int len;
10 }pre[50080];
11 vector<pair<int ,int > >query[50050];
12 vector<pair <int ,int > >vec[50100];
13 int ans[50050],vis[50100];
14 int Find(int x)
15 {
16     if(pre[x].f==x)
17     {
18         return x;
19     }
20     else
21     {
22         pre[x].f=Find(pre[x].f);
23         return pre[x].f;
24     }
25 }
26 int dfs(int u,int fa,int length)
27 {
28     pre[u].f=u;
29     pre[u].len=pre[fa].len+length;
30     vis[u]=1;
31     for(int i=0;i<vec[u].size();i++)
32     {
33         int v=vec[u][i].first;
34         int l=vec[u][i].second;
35         if(v==fa)   continue;
36         dfs(v,u,l);
37     }
38     for(int i=0;i<query[u].size();i++)
39     {
40         int v=query[u][i].first;
41         int id=query[u][i].second;
42         if(vis[v]==1)
43         {
44              ans[id]=pre[v].len+pre[u].len-2*pre[Find(v)].len;
45         }
46     }
47     pre[u].f=fa;
48 }
49 int main()
50 {
51     int n,m,x,y,l;
52     char c;
53     while(scanf("%d%d",&n,&m)!=EOF)
54     {
55         memset(vis,0,sizeof(vis));
56         for(int i=0;i<m;i++)
57         {
58             scanf("%d%d%d %c",&x,&y,&l,&c);
59             vec[x].push_back({y,l});
60             vec[y].push_back({x,l});
61             vis[y]=1;
62         }
63         int k;
64         cin>>k;
65         for(int i=0;i<k;i++)
66         {
67             scanf("%d%d",&x,&y);
68             query[x].push_back({y,i});
69             query[y].push_back({x,i});
70         }
71         for(int i=1;i<=n;i++)
72         {
73             if(vis[i]==0)
74             {
75                 memset(vis,0,sizeof(vis));
76                 dfs(i,-1,0);
77                 break;
78             }
79         }
80         for(int i=0;i<k;i++)
81             cout<<ans[i]<<endl;
82         memset(ans,0,sizeof(ans));
83         memset(pre,0,sizeof(pre));
84         for(int i=0;i<=n;i++)
85         {
86             vec[i].clear();
87             query[i].clear();
88         }
89     }
90     return 0;
91 }