Codeforces Round #482 (Div. 2) C Kuro and Walking Route
Kuro is living in a country called Uberland, consisting of nn towns, numbered from 11 to nn, and n?1n?1 bidirectional roads connecting these towns. It is possible to reach each town from any other. Each road connects two towns
Oddly, there are 2 special towns in Uberland named Flowrisa (denoted with the index
Kuro wants to know how many pair of city (u,v) he can take as his route. Since he’s not really bright, he asked you to help him with this problem.
InputThe first line contains three integers n, x and y (1≤n≤3?105,1≤x,y≤n1≤n≤3?105,1≤x,y≤n, x≠y) - the number of towns, index of the town Flowrisa and index of the town Beetopia, respectively.
n?1lines follow, each line contains two integers a and b (1≤a,b≤n1≤a,b≤n, a≠b), describes a road connecting two towns a and b.
It is guaranteed that from each town, we can reach every other town in the city using the given roads. That is, the given map of towns and roads is a tree.
OutputA single integer resembles the number of pair of towns (u,v) that Kuro can use as his walking route.
Examples input Copy3 1 3output Copy
1 2
2 3
5input Copy
3 1 3output Copy
1 2
1 3
4Note
On the first example, Kuro can choose these pairs:
- (1,2)(1,2): his route would be 1→2,
- (2,3)(2,3): his route would be 2→3,
- (3,2)(3,2): his route would be 3→2
- (2,1)(2,1): his route would be 2→1,
- (3,1)(3,1): his route would be 3→2→1.
Kuro can‘t choose pair (1,3)(1,3) since his walking route would be 1→2→31→2→3, in which Kuro visits town 11 (Flowrisa) and then visits town 33(Beetopia), which is not allowed (note that pair (3,1)(3,1) is still allowed because although Kuro visited Flowrisa and Beetopia, he did not visit them in that order).
On the second example, Kuro can choose the following pairs:
- (1,2)(1,2): his route would be 1→2,
- (2,1)(2,1): his route would be 2→1,
- (3,2)(3,2): his route would be 3→1→2,
- (3,1)(3,1): his route would be 3→1.
題意 給出一顆n個定點的樹 樹上有兩個點 想 x,y 任意兩點相互可達 但u到v的路徑上先經過x再經過y是不允許的 (u,v)和(v , u)是兩條不相同的路徑
問滿足上述條件的路徑有多少條
解析
我們選定一個點作為根節點1 size[ i ]表示 以i為根節點的子樹大小
若x的祖先沒有y y 的祖先沒有x 那麽答案就是size[x]*size[y]
若x 的祖先有y 那麽我們要找x的祖先 且 是y的兒子的那個節點 fa 答案就是(n-size[fa])*size[x]
同理得到另一種情況
AC代碼
#include<bits/stdc++.h> using namespace std; const int maxn = 3e5+50 ,mod = 998244353,inf=0x3f3f3f3f; const double pi=acos(-1.0); typedef long long ll; vector<int> g[maxn]; ll siz[maxn],fa[maxn]; ll n,x,y; int vis[maxn]; void dfs1(int k,int f) { vis[k]=1; siz[k]=1; fa[k]=f; for(int i=0;i<g[k].size();i++) { int u=g[k][i]; if(vis[u]==0&&u!=f) { dfs1(u,k); siz[k]+=siz[u]; } } } int dfs2(int k1,int k2) { if(fa[k1]==k2) return k1; else if(fa[k1]==0) return 1; else return dfs2(fa[k1],k2); } int main() { cin>>n>>x>>y; for(int i=0;i<n-1;i++) { int u,v; cin>>u>>v; g[u].push_back(v); g[v].push_back(u); } memset(siz,0,sizeof(siz)); memset(vis,0,sizeof(vis)); dfs1(1,0); ll ans; int f1=dfs2(x,y),f2=dfs2(y,x); //cout<<f1<<" "<<f2<<endl; if(f1==f2) ans=siz[x]*siz[y]; else if(f1==1) ans=(n-siz[f2])*siz[y]; else ans=(n-siz[f1])*siz[x]; // for(int i=1;i<=n;i++) // cout<<i<<" "<<fa[i]<<endl; cout<<n*(n-1)-ans<<endl; }
Codeforces Round #482 (Div. 2) C Kuro and Walking Route