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  • POJ 1986：Distance Queries

    Distance Queries

    Time Limit: 2000MS   Memory Limit: 30000K
    Total Submissions: 18139   Accepted: 6248
    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.

    题意

    n个点，m条边，每两个相连的点有一个距离，对于每次询问，求出u，v的距离

    思路

    因为题中给出的图是一个树（Navigation Nightmare题目链接：http://poj.org/problem?id=1984）

    对于树上的两点距离，我们有：dis(u,v)=dis(u,root)+dis(v,toot)-2*dis(lca(u,v),root)

    预处理出来每个点到根节点的距离，在查询的时候求出u，v两点的lca，然后利用上述公式计算即可

    因为是一棵树，所以可以以任意一个节点作为根节点

    代码

      1 #include <algorithm>
  2 #include <iostream>
  3 #include <string.h>
  4 #define ll long long
  5 #define ull unsigned long long
  6 #define ms(a,b) memset(a,b,sizeof(a))
  7 const int inf=0x3f3f3f3f;
  8 const ll INF=0x3f3f3f3f3f3f3f3f;
  9 const int maxn=2e5+10;
 10 const int mod=1e9+7;
 11 const int maxm=1e3+10;
 12 using namespace std;
 13 int f[maxn];
 14 int find(int x)
 15 {
 16     if(f[x]!=x)
 17         f[x]=find(f[x]);
 18     return f[x];
 19 }
 20 inline void join(int x,int y)
 21 {
 22     int dx=f[x],dy=f[y];
 23     if(dx!=dy)
 24         f[dy]=dx;
 25 }
 26 struct Edge
 27 {
 28     int to,Next;
 29     int value;
 30 }edge[maxn];
 31 int tot1;
 32 int head1[maxn];
 33 inline void add_edge(int u,int v,int w)
 34 {
 35     edge[tot1].to=v;
 36     edge[tot1].value=w;
 37     edge[tot1].Next=head1[u];
 38     head1[u]=tot1++;
 39 }
 40 int vist[maxn];
 41 int dis[maxn];
 42 // 预处理每个点到根节点的距离
 43 void dfs(int u,int len)
 44 {
 45     dis[u]=len;
 46     vist[u]=1;
 47     for(int i=head1[u];~i;i=edge[i].Next)
 48     {
 49         int v=edge[i].to;
 50         if(!vist[v])
 51             dfs(v,len+edge[i].value);
 52     }
 53 }
 54 struct Query
 55 {
 56     int to,nex;
 57     int index;
 58 }query[maxn];
 59 int head2[maxn];
 60 int tot2;
 61 inline void add_query(int u,int v,int index)
 62 {
 63     query[tot2].to=v;
 64     query[tot2].index=index;
 65     query[tot2].nex=head2[u];
 66     head2[u]=tot2++;
 67     query[tot2].to=u;
 68     query[tot2].index=index;
 69     query[tot2].nex=head2[v];
 70     head2[v]=tot2++;
 71 }
 72 int vis[maxn];
 73 int fa[maxn];
 74 int ans[maxn];
 75 void LCA(int u)
 76 {
 77     fa[u]=u;
 78     vis[u]=1;
 79     for(int i=head1[u];~i;i=edge[i].Next)
 80     {
 81         int v=edge[i].to;
 82         if(vis[v])
 83             continue;
 84         LCA(v);
 85         join(u,v);
 86         fa[find(u)]=u;
 87     }
 88     for(int i=head2[u];~i;i=query[i].nex)
 89     {
 90         int v=query[i].to;
 91         if(vis[v])
 92             ans[query[i].index]=fa[find(v)];
 93     }
 94 }
 95 inline void init(int n)
 96 {
 97     tot1=tot2=0;
 98     ms(head1,-1);
 99     ms(head2,-1);
100     ms(vis,0);
101     ms(vist,0);
102     ms(fa,0);
103     ms(dis,0);
104     for(int i=1;i<=n;i++)
105         f[i]=i;
106 }
107 int x[maxn],y[maxn];
108 int main(int argc, char const *argv[])
109 {
110     #ifndef ONLINE_JUDGE
111         freopen("/home/wzy/in.txt", "r", stdin);
112         freopen("/home/wzy/out.txt", "w", stdout);
113         srand((unsigned int)time(NULL));
114     #endif
115     ios::sync_with_stdio(false);
116     cin.tie(0);
117     int n,m,q;
118     while(cin>>n>>m)
119     {
120         init(n);
121         int u,v,w;
122         char ch[3];
123         for(int i=0;i<m;i++)
124             cin>>u>>v>>w>>ch,add_edge(u,v,w),add_edge(v,u,w);
125         dfs(1,0);
126         cin>>q;
127         for(int i=0;i<q;i++)
128             cin>>x[i]>>y[i],add_query(x[i],y[i],i);
129         LCA(1);
130         for(int i=0;i<q;i++)
131             cout<<dis[x[i]]+dis[y[i]]-2*dis[ans[i]]<<endl;
132     }
133     #ifndef ONLINE_JUDGE
134         cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
135     #endif
136     return 0;
137 }
    View Code
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  • 原文地址：https://www.cnblogs.com/Friends-A/p/11303707.html
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