zoj 3949 Edge to the Root 2017浙大校赛B
Given a tree with n vertices, we want to add an edge between vertex 1 and vertex x, so that the sum of d(1, v) for all vertices v in the tree is minimized, where d(u, v) is the minimum number of edges needed to pass from vertex u to vertex v. Do you know which vertex x we should choose?
Recall that a tree is an undirected connected graph with n vertices and n - 1 edges.
Input
There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
The first line contains an integer n (1 ≤ n ≤ 2 × 105), indicating the number of vertices in the tree.
Each of the following n - 1 lines contains two integers u and v (1 ≤ u, v ≤ n), indicating that there is an edge between vertex u and v in the tree.
It is guaranteed that the given graph is a tree, and the sum of n over all test cases does not exceed 5 × 105. As the stack space of the online judge system is not very large, the maximum depth of the input tree is limited to about 3 × 104.
We kindly remind you that this problem contains large I/O file, so it's recommended to use a faster I/O method. For example, you can use scanf/printf instead of cin/cout in C++.
Output
For each test case, output a single integer indicating the minimum sum of d(1, v) for all vertices v in the tree (NOT the vertex x you choose).
Sample Input
2 6 1 2 2 3 3 4 3 5 3 6 3 1 2 2 3
Sample Output
8 2
Hint
For the first test case, if we choose x = 3, we will have
d(1, 1) + d(1, 2) + d(1, 3) + d(1, 4) + d(1, 5) + d(1, 6) = 0 + 1 + 1 + 2 + 2 + 2 = 8
It's easy to prove that this is the smallest sum we can achieve.
现场赛的时候应该是写出正解了...但是不知道为什么挂了
结果回来按照记忆默了一遍七分钟直接就1A了
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枚举每个点，他子树中所有节点都会经过他到根，他到根上中点这段路径上的所有节点和他们的子树也会经过他到根
那么考虑按照深度枚举每个转移点，假设现在是d,中点为x，下一个枚举位置为t,son[]表示子树中节点个数
显然当枚举点往下移的时候，t所有的子节点到根的路径都会从由d跳转变为由t跳转，这样整体会减一
然后不考虑x的下移，那么x到d这段路径上所有点和他们子树中的节点的路径都会加1
再考虑x是否下移，移动的话则将x不包括t的所有子树中的点直接走向根，这里的贡献和到d再到根是一致的，因此不需要变化。
最后通过回溯来消除变化，这样直接将所有点dfs一遍即可求出最小的答案
#include
#include
#include
#include
using namespace std;
struct line
{
int s,t;
int next;
}a[400001];
int head[200001];
int edge;
inline void add(int s,int t)
{
a[edge].next=head[s];
head[s]=edge;
a[edge].s=s;
a[edge].t=t;
}
int fa[200001],son[200001],dep[200001];
inline void dfs1(int d)
{
int i;
for(i=head[d];i!=0;i=a[i].next)
{
int t=a[i].t;
if(t!=fa[d])
{
fa[t]=d;
dep[t]=dep[d]+1;
dfs1(t);
son[d]+=son[t]+1;
}
}
}
long long sx,ans,sx1,sx2;
int nxt[200001];
inline void dfs2(int d,int x)
{
// printf("d:%d x:%d sx:%lld sx1:%lld sx2:%lld\n",d,x,sx,sx1,sx2);
int i;
for(i=head[d];i!=0;i=a[i].next)
{
int t=a[i].t;
if(t!=fa[d])
{
long long lx=x,lsx=sx,lsx1=sx1,lsx2=sx2;
if(dep[x]1)
sx-=sx1;
sx+=sx2;
ans=min(ans,sx);
nxt[d]=t;
dfs2(t,x);
x=lx;
sx=lsx;
sx1=lsx1;
sx2=lsx2;
}
}
}
int main()
{
int T;
scanf("%d",&T);
while(T>0)
{
T--;
int n;
scanf("%d",&n);
edge=0;
memset(fa,0,sizeof(fa));
memset(son,0,sizeof(son));
memset(head,0,sizeof(head));
int i,s,t;
for(i=1;i<=n-1;i++)
{
scanf("%d%d",&s,&t);
edge++;
add(s,t);
edge++;
add(t,s);
}
dep[1]=0;
dfs1(1);
ans=0;
for(i=1;i<=n;i++)
ans+=dep[i];
sx=ans;
sx1=son[1]+1;
sx2=0;
dfs2(1,1);
printf("%lld\n",ans);
}
return 0;
}