Poj 1986 distance queries LCA

Question link: http://poj.org/problem? Id = 1986

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 I Ntegers, 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!

Description: The distance between n (n <= 40000) farms and some farms is known. There are many groups of inquiries online asking about the distance between each farm.

Algorithm Analysis: online and offline algorithms using LCA can be used for this question.

Offline LCA idea: Use a farm as the root of the tree and use an array to save the distance from each farm to the root of the tree as Dist [u]. then the distance between the farm U and V is = DIST [u] + dist [v]-2 * Dist [Q] (Q is LCA (u, v )).

Although my practice can be AC, It is very slow and there is a lot of room for optimization: while the two nodes keep approaching the root of the tree, the sum of the weights (distance between farms) on each path that goes up one layer is the answer. I used the offline LCA code to apply it to the online method. I thought about how to find the DIST array in the offline thinking when finding the depth of each node, which is simpler and faster.

  1 #include<iostream>  2 #include<cstdio>  3 #include<cstring>  4 #include<cstdlib>  5 #include<cmath>  6 #include<algorithm>  7 #include<vector>  8 #define inf 0x7fffffff  9 using namespace std; 10 const int maxn=40000+10; 11 const int max_log_maxn=16; 12  13 int n,m,root; 14 vector<pair<int,int> > G[maxn]; 15 int father[max_log_maxn][maxn],d[maxn]; 16 int sum[max_log_maxn][maxn]; 17  18 void dfs(int u,int p,int depth) 19 { 20     father[0][u]=p; 21     d[u]=depth; 22     int num=G[u].size(); 23     for (int i=0 ;i<num ;i++) 24     { 25         int v=G[u][i].first; 26         int len=G[u][i].second; 27         if (v != p) 28         { 29             sum[0][v]=len; 30             dfs(v,u,depth+1); 31         } 32     } 33 } 34  35 void init() 36 { 37     dfs(root,-1,0); 38     for (int k=0 ;k+1<max_log_maxn ;k++) 39     { 40         for (int i=1 ;i<=n ;i++) 41         { 42             if (father[k][i]<0) 43             { 44                 father[k+1][i]=-1; 45                 sum[k+1][i]=0; 46             } 47             else 48             { 49                 father[k+1][i]=father[k][father[k][i] ]; 50                 sum[k+1][i]=sum[k][i]+sum[k][father[k][i] ]; 51             } 52         } 53     } 54 } 55  56 int LCA(int u,int v) 57 { 58     if (d[u]>d[v]) swap(u,v); 59     int Sum=0; 60     for (int k=0 ;k<max_log_maxn ;k++) 61     { 62         if ((d[v]-d[u])>>k & 1) 63         { 64             Sum += sum[k][v]; 65             v=father[k][v]; 66         } 67     } 68     if (u==v) return Sum; 69     for (int k=max_log_maxn-1 ;k>=0 ;k--) 70     { 71         if (father[k][u] != father[k][v]) 72         { 73             Sum += sum[k][u]; 74             Sum += sum[k][v]; 75             u=father[k][u]; 76             v=father[k][v]; 77         } 78     } 79     Sum += sum[0][u]; 80     Sum += sum[0][v]; 81     return Sum; 82 } 83  84 int main() 85 { 86     while (scanf("%d%d",&n,&m)!=EOF) 87     { 88         int u,v,length; 89         char ch[2]; 90         for (int i=0 ;i<=n ;i++) G[i].clear(); 91         for (int i=0 ;i<max_log_maxn ;i++) 92         { 93             for (int j=0 ;j<maxn ;j++) 94                 sum[i][j]=0; 95         } 96         for (int i=0 ;i<m ;i++) 97         { 98             scanf("%d%d%d%s",&u,&v,&length,ch); 99             G[u].push_back(make_pair(v,length));100             G[v].push_back(make_pair(u,length));101         }102         root=1;103         init();104         int k;105         scanf("%d",&k);106         for (int i=0 ;i<k ;i++)107         {108             scanf("%d%d",&u,&v);109             printf("%d\n",LCA(u,v));110         }111     }112     return 0;113 }

 

Poj 1986 distance queries LCA