Problem Description
Give you a tree with N vertices and N‐ 1 edges, and then ask you Q queries on “which vertex is Y's son that has the smallest number and which vertex is Y’s descendants that has the smallest number if we choose X as the root of the entire tree?”
Input
The first line of input is an integer T (T<=10) means the case number.
The first line of each test case contains N(2 ≤ N ≤ 100,000) and Q(1 ≤ Q ≤ 100,000).
Each of the following N ‐ 1 lines of the test case contains two integers a(1 ≤ a ≤ N) and b(1 ≤ b ≤ N) indicating an edge between a and b.
Each of the following Q lines of the test case contains two integers X(1 ≤ X ≤ N) and Y(1 ≤ Y ≤ N, Y ≠ X) indicating an query.
Output
For each query, output the Y's son which has the smallest number and Y's descendant that has the smallest number if X is the root of the entire tree. If Y has no sons then output “no answers!”. There is an empty line after each case.
Sample Input
1
7 3
1 2
1 5
2 3
2 4
5 6
5 7
1 2
5 3
3 2
Sample Output
3 3
no answers!
1 1
给出一棵树,根节点是不确定的,每个节点有一个不同标号,标号在1到n之间。
然后问当x为根的时候y节点的儿子中的标号最小值和所有子孙节点中的最小值。
首先可以以节点1为根进行一次预处理,求出每个节点的儿子中的最
小值和次小值以及子孙节点中的最小值和次小值。同时记录一个dfs序列,可以
o(1)的判断两个节点是不是包含关系。
接下来分类讨论,读入x,y。
1.如果y不包含x的,那么直接输出y儿子中的最小值和子孙中的最小值即可(注意判断是否有儿子)。
2.如果y包含了x
(1)如果y不是节点1,那么找到x到y路径中最接近y的点z
若z是y儿子中的最小值,输出y儿子的次小值和1.
否则输出y儿子中的最小值和1;
(2)如果y是节点1,且y没有其他分支输出no answer!
#include<stdio.h>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn = 100005;
vector<int> t, tree[maxn];
int tot, T, N, Q, x, y, m[maxn][2], M[maxn][2], dfn[maxn][2], fa[maxn];
void begin()
{
tot = 0;
for (int i = 1; i <= N; i++)
{
tree[i].clear();
fa[i] = 0;
m[i][0] = m[i][1] = 0x7FFFFFFF;
M[i][0] = M[i][1] = 0x7FFFFFFF;
dfn[i][0] = dfn[i][1] = 0;
}
}
void dfs(int x, int xx)
{
fa[x] = xx; dfn[x][0] = ++tot; t.clear();
for (int i = 0; i < tree[x].size(); i++)
if (tree[x][i] != xx) t.push_back(tree[x][i]);
tree[x] = t;
for (int i = 0; i < tree[x].size(); i++)
{
int y = tree[x][i];
dfs(y, x);
if (y < m[x][0]) m[x][1] = m[x][0], m[x][0] = y;
else m[x][1] = min(m[x][1], y);
y = min(y, M[y][0]);
if (y < M[x][0]) M[x][1] = M[x][0], M[x][0] = y;
else M[x][1] = min(M[x][1], y);
}
dfn[x][1] = ++tot;
}
int half(int x, int y)
{
int q = 0, h = tree[y].size() - 1, mid;
while (true)
{
mid = (q + h) >> 1;
if (dfn[x][0] >= dfn[tree[y][mid]][0] && dfn[x][1] <= dfn[tree[y][mid]][1]) return tree[y][mid];
if (dfn[x][0] < dfn[tree[y][mid]][0]) h = mid - 1; else q = mid + 1;
}
}
int main()
{
scanf("%d", &T);
while (T--)
{
scanf("%d%d", &N, &Q);
begin();
for (int i = 1; i < N; i++)
{
scanf("%d%d", &x, &y);
tree[x].push_back(y);
tree[y].push_back(x);
}
dfs(1, 1);
while (Q--)
{
scanf("%d%d", &x, &y);
if (dfn[x][0]>dfn[y][0] && dfn[x][1] < dfn[y][1])
{
int z = half(x, y);
if (y != 1)
{
if (z == m[y][0]) printf("%d 1\n", min(m[y][1], fa[y]));
else printf("%d 1\n", min(m[y][0], fa[y]));
}
else
{
if (tree[y].size() == 1) printf("no answers!\n");
else
{
if (z == m[y][0]) printf("%d ", m[y][1]);
else printf("%d ", m[y][0]);
if (min(z, M[z][0]) == M[y][0]) printf("%d\n", M[y][1]);
else printf("%d\n", M[y][0]);
}
}
}
else
if (dfn[y][0] + 1 == dfn[y][1]) printf("no answers!\n");
else printf("%d %d\n", m[y][0], M[y][0]);
}
printf("\n");
}
}