H. Mad City Codeforces Round 898 (Div. 4)

Problem - H - Codeforces

题目大意：有一个n个点的图，有n条边，西安有两个人A，B分别位于点a/b，每轮两个人同时开始向相邻点移动，问B有没有可能永远不和A汇合

3<=n<=2e5;1<=a,b<=n

思路：如果要永远不汇合，那么他们最终肯定在一个环里，而边的数量等于点数，说明图中有且只有一个环，那么B只要比A先到这个环即可。

所以我们先跑一个dfs找到所有环上的点，然后再以b为起点，找到b距离环上的哪个点最近，且是哪个点，然后再从a出发跑bfs找到b到那个点的距离，如果b一开始就在环上且到那个点比a更近，就有必胜策略，同时如果ab重合一定没有必胜策略

//#include<__msvc_all_public_headers.hpp>
#include
using namespace std;
typedef long long ll;
const ll MOD = 1e9 + 7;
const int N = 2e5 + 5;
int head[N];
int cnt = 0;
int n;
struct Edge
{
	int v, next;
}e[N * 2];
void addedge(int u, int v)
{
	e[++cnt].v = v;
	e[cnt].next = head[u];
	head[u] = cnt;
}
bool vis[N];
bool incycle[N];
bool flag = 0;
void init()
{
	cnt = 0;
	for (int i = 1; i <= n; i++)
	{
		flag = 0;
		head[i] = -1;
		vis[i] = 0;
		incycle[i] = 0;
	}
}
int path[N];
void find_cycle(int u, int fa)
{//标记哪些点在环上
	vis[u] = 1;
	for (int i = head[u]; ~i; i = e[i].next)
	{
		int v = e[i].v;
		if (v == fa || flag)
			continue;
		if (vis[v])
		{//出现访问过的点，说明这里成环
			flag = 1;
			int now = u;
			while (now != v)
			{
				incycle[now] = 1;
				now = path[now];
			}
			incycle[now] = 1;
			return;
		}
		path[v] = u;//记录父节点
		find_cycle(v, u);
	}
}
void solve()
{
	int a, b;
	cin >> n >> a >> b;
	init();
	for (int i = 1; i <= n; i++)
	{
		int u, v;
		cin >> u >> v;
		addedge(u, v);
		addedge(v, u);
	}
	find_cycle(1, 0);
	queue>q;
	q.push({ b,0 });
	int dis1 = -1;
	int poi;
	for(int i = 1; i <= n; i++)
	{
		vis[i] = 0;
	}
	vis[b] = 1;
	while (!q.empty())
	{//以b为起点
		int u = q.front().first;
		int d = q.front().second;
		if (incycle[u])
		{//找到最近的在环上的点
			poi = u;
			dis1 = d;
			break;
		}
		q.pop();
		for (int i = head[u]; ~i; i = e[i].next)
		{
			int v = e[i].v;
			if (vis[v])
				continue;
			vis[v] = 1;
			q.push({ v, d + 1 });
		}
	}
	queue>q2;
	q2.push({ a,0 });
	int dis2 = -1;
	for (int i = 1; i <= n; i++)
	{
		vis[i] = 0;
	}
	vis[a] = 1;
	while (!q2.empty())
	{//以a为起点
		int u = q2.front().first;
		int d = q2.front().second;
		if (u == poi)
		{//找到b要去的环上的点
			dis2 = d;
			break;
		}
		q2.pop();
		for (int i = head[u]; ~i; i = e[i].next)
		{
			int v = e[i].v;
			if (vis[v])
				continue;
			vis[v] = 1;
			q2.push({ v, d + 1 });
		}
	}
	cout << ((( dis1 == 0 || dis2 > dis1) && a != b) ? "YES" : "NO") << endl;//b先到且b和a起始不重合
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int t;
	cin >> t;
	while (t--)
	{
		solve();
	}
	return 0;
}