【题解】codeforces1029E[Codeforces Round #506 (Div. 3)]E.Tree with Small Distances 优先队列+dfs
题目链接
Description
You are given an undirected tree consisting of n n n vertices. An undirected tree is a connected undirected graph with n − 1 n−1 n−1 edges.
Your task is to add the minimum number of edges in such a way that the length of the shortest path from the vertex 1 1 1 to any other vertex is at most 2 2 2. Note that you are not allowed to add loops and multiple edges.
Input
The first line contains one integer n n n ( 2 ≤ n ≤ 2 × 1 0 5 ) (2≤n≤2\times10^5) (2≤n≤2×105) — the number of vertices in the tree.
The following n − 1 n−1 n−1 lines contain edges: edge i i i is given as a pair of vertices u i , v i u_i,v_i ui,vi ( 1 ≤ u i , v i ≤ n ) (1≤ui,vi≤n) (1≤ui,vi≤n). It is guaranteed that the given edges form a tree. It is guaranteed that there are no loops and multiple edges in the given edges.
Output
Print a single integer — the minimum number of edges you have to add in order to make the shortest distance from the vertex 1 1 1 to any other vertex at most 2 2 2. Note that you are not allowed to add loops and multiple edges.
Examples
Input
7
1 2
2 3
2 4
4 5
4 6
5 7
Output
2
Input
7
1 2
1 3
2 4
2 5
3 6
1 7
Output
0
Input
7
1 2
2 3
3 4
3 5
3 6
3 7
Output
1
Note
The tree corresponding to the first example:
The answer is 2 2 2, some of the possible answers are the following: [ ( 1 , 5 ) , ( 1 , 6 ) ] , [ ( 1 , 4 ) , ( 1 , 7 ) ] , [ ( 1 , 6 ) , ( 1 , 7 ) ] [(1,5),(1,6)], [(1,4),(1,7)], [(1,6),(1,7)] [(1,5),(1,6)],[(1,4),(1,7)],[(1,6),(1,7)].
The tree corresponding to the second example:
The answer is 0 0 0.
The tree corresponding to the third example:
The answer is 1 1 1, only one possible way to reach it is to add the edge ( 1 , 3 ) (1,3) (1,3).
要求连最少的边使以 1 1 1 为根的树上所有子节点深度不超过 2 2 2 。 d f s dfs dfs 一遍求出每个点的深度，把每个深度超过 2 2 2 的点加入堆中。每次取出堆顶节点，如果没有被更新，就向他的父节点连一条返祖边，然后遍历与父节点直接相连的节点标记。
#include
#include
#include
using namespace std;
const int N=2e5+10;
int dep[N],fa[N],n,hd[N],tot,vis[N];
struct Edge{
int v,nx;
}e[N<<1];
void add(int u,int v)
{
e[tot].v=v;
e[tot].nx=hd[u];
hd[u]=tot++;
}
void dfs(int u,int f)
{
for(int i=hd[u];~i;i=e[i].nx)
{
int v=e[i].v;
if(v==f)continue;
fa[v]=u;dep[v]=dep[u]+1;
dfs(v,u);
}
}
int main()
{
//freopen("in.txt","r",stdin);
memset(hd,-1,sizeof(hd));
scanf("%d",&n);
int u,v,ans=0;
for(int i=1;i >q;
for(int i=1;i<=n;i++)if(dep[i]>2)q.push(make_pair(dep[i],i));
while(q.size())
{
u=q.top().second;q.pop();
if(vis[u])continue;
ans++;vis[fa[u]]=1;
for(int i=hd[fa[u]];~i;i=e[i].nx)vis[e[i].v]=1;
}
printf("%d\n",ans);
return 0;
}
总结
可能可以证明每次去最深的节点是最优的，反例是容易举出的。