题目链接:CF1405D「Tree Tag」 。
Alice and Bob are playing a fun game of tree tag.
The game is played on a tree of n vertices numbered from 1 to n. Recall that a tree on n vertices is an undirected, connected graph with n-1 edges.
Initially, Alice is located at vertex a, and Bob at vertex b. They take turns alternately, and Alice makes the first move. In a move, Alice can jump to a vertex with distance at most dadada from the current vertex. And in a move, Bob can jump to a vertex with distance at most dbfrom the current vertex. The distance between two vertices is defined as the number of edges on the unique simple path between them. In particular, either player is allowed to stay at the same vertex in a move. Note that when performing a move, a player only occupies the starting and ending vertices of their move, not the vertices between them.
If after at most 10^{100} moves, Alice and Bob occupy the same vertex, then Alice is declared the winner. Otherwise, Bob wins.
Determine the winner if both players play optimally.
Each test contains multiple test cases. The first line contains the number of test cases t (1 \leq t \leq 10^4). Description of the test cases follows.
The first line of each test case contains five integers n,a,b,da,db(2 \leq n \leq 10^5; 1 \leq a,b \leq n; a \neq b; 1 \leq da,db \leq n-1) — the number of vertices, Alice’s vertex, Bob’s vertex, Alice’s maximum jumping distance, and Bob’s maximum jumping distance, respectively.
The following n-1 lines describe the edges of the tree. The i-th of these lines contains two integers u, v (1 \leq u, v \leq n, u \neq v), denoting an edge between vertices u and v. It is guaranteed that these edges form a tree structure.
It is guaranteed that the sum of n across all test cases does not exceed 10^5.
For each test case, output a single line containing the winner of the game: “Alice” or “Bob”.
4
4 3 2 1 2
1 2
1 3
1 4
6 6 1 2 5
1 2
6 5
2 3
3 4
4 5
9 3 9 2 5
1 2
1 6
1 9
1 3
9 5
7 9
4 8
4 3
11 8 11 3 3
1 2
11 9
4 9
6 5
2 10
3 2
5 9
8 3
7 4
7 10
Alice
Bob
Alice
Alice
题目描述感觉有点诘屈聱牙,这个问题主要就是说如果在有限步骤内,Alice 和 Bob 能够同时移动到同一个顶点上,那么 Alice 赢;否则 Bob 赢。就相当于 Bob 逃,Alice 追,追得上 Alice 赢,追不上 Bob 赢。这道题蒟蒻的我也是看了官方题解才弄懂的(Orz)。
对于这道题应该分情况讨论:设树中 a 到 b 的唯一简单路径长度为 (a,b),树的直径为 diameterdiameterdiameter,则
#include <bits/stdc++.h>
#define ll int
#define MAXN 100005
using namespace std;
// 前向星存边
ll cnt;
ll head[MAXN];
struct edge{
ll to, next;
}e[2*MAXN];
void init() {
cnt = 0;
memset(head, -1, sizeof(head));
}
void addEdge(ll u, ll v) {
e[cnt].next = head[u];
e[cnt].to = v;
head[u] = cnt++;
e[cnt].next = head[v];
e[cnt].to = u;
head[v] = cnt++;
}
ll dia;
ll vis[MAXN];
ll depth[MAXN];
vector <ll> v[MAXN];
// dfs 求出树的直径
// 同时求出 a 和 b 的距离
ll dfs(ll a, ll b) {
vis[a] = 1;
ll res = 0;
v[a].clear();
for(ll i = head[a]; i != -1; i = e[i].next) {
ll u = e[i].to;
if(!vis[u]) {
depth[u] = depth[a] + 1;
res = dfs(u, b);
v[a].push_back(res+1);
}
}
if(v[a].size()) {
sort(v[a].begin(),v[a].end(), greater<ll>());
if(v[a].size() > 1) {
dia = max(dia, v[a][0]+v[a][1]);
} else {
dia = max(dia, v[a][0]);
}
res = v[a][0];
}
return res;
}
int main()
{
ll t;
scanf("%d", &t);
for(ll i = 0; i < t; ++i) {
ll n, a, b, da, db;
init();
scanf("%d%d%d%d%d", &n, &a, &b, &da, &db);
for(ll j = 0; j < n-1; ++j) {
ll u, v;
scanf("%d%d", &u, &v);
addEdge(u, v);
}
bool mark;
dia = 0;
memset(vis, 0, sizeof(vis));
memset(depth, 0, sizeof(depth));
dfs(a, b);
ll dist = depth[b];
if(dist <= da) {
mark = true;
} else if(dia <= 2*da) {
mark = true;
} else if(2*da < db) {
mark = false;
} else {
mark = true;
}
printf("%s\n", mark ? "Alice" : "Bob");
}
return 0;
}