Distance in Tree CodeForces, distancecodeforces

Source: Internet
Author: User

Distance in Tree CodeForces, distancecodeforces

Distance in Tree CodeForces-161D

For a tree with n nodes, the distance between any two points is 1. Now it is a bit u, v, and the shortest distance between u and v is k, calculate the number of such points (u, v) (u, v)/(v, u) as a pair ).

Method:

Ans [I] [k] indicates the number of subnodes whose distance from the I node is k.

Ans [I) [k] = sum {ans [son] [k-1]}

Ans [I] [0] = 1

Sum [I] indicates that (u, v) is the subnode of I and the shortest path of (u, v) is over I.

Sum [I] = sum {ans [I] [p] * ans [I] [k-p]} // incorrect.

Sum [I] = sum {ans [son] [p] * sum {ans [otherson] [k-p-2]} // pair, but too slow

Sum [I] = sum {ans [son] [p] * (ans [I] [k-p-1]-ans [son] [k-p-2])} // rewrite (unexpected)

Because (u, v)/(v, u) is a pair, the actual I-point-related answer (that is, an endpoint is I-point, or (u, v) and (u, v) Shortest Path over I point) is:

$ Ans [I] [k] + sum [I]/2 $

Because you do not need to separate the points, you can directly add these values with an ans without opening sum.

 1 #include<cstdio> 2 #include<vector> 3 using namespace std; 4 typedef long long LL; 5 struct Edge 6 { 7     LL to,next; 8 }edge[100001]; 9 LL first1[50010],num_edge;10 LL ans[50001][501];11 LL ans1,ans2,n,k2;12 bool vis[50001];13 void dfs(LL u)14 {15     LL k=first1[u],i,j;16     vector<LL> temp;17     ans[u][0]=1;18     vis[u]=true;19     while(k!=0)20     {21         LL &v=edge[k].to;22         if(!vis[v])23         {24             dfs(v);25             for(i=1;i<=k2;i++)26                 ans[u][i]+=ans[v][i-1];27             temp.push_back(v);28         }29         k=edge[k].next;30     }31     ans2+=ans[u][k2];32     for(i=0;i<temp.size();i++)33         for(j=0;j<=k2-2;j++)34             ans1+=ans[temp[i]][j]*(ans[u][k2-j-1]-ans[temp[i]][k2-j-2]);35 }36 int main()37 {38     LL i,a,b;39     scanf("%I64d%I64d",&n,&k2);40     for(i=1;i<n;i++)41     {42         scanf("%I64d%I64d",&a,&b);43         edge[++num_edge].to=b;44         edge[num_edge].next=first1[a];45         first1[a]=num_edge;46         edge[++num_edge].to=a;47         edge[num_edge].next=first1[b];48         first1[b]=num_edge;49     }50     dfs(1);51     printf("%I64d",ans2+ans1/2);52     return 0;53 }

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