P8025 [ONTAK2015] Związek Harcerstwa Bajtockiego

给一个 Dijkstra 的 $48$ 分算法，正解想不出来了？

思路倒也很简单，Dijkstra 过程中记录一下路径，就可以跑最短路了，如果最短路小于等于 $t$，直接转移。不然就到走 $t$ 步的位置上。

代码：

#include <iostream>
#include <cstdio>
#include <queue>
#include <vector>
#include <cstring>
using namespace std;

#define int long long

const int N = 1e6 + 5;

vector<int> G[N];
int n, m, k, now, pre[N], dis[N];
bool vis[N];
int d, t;

struct Node
{
	int place, dist;
	Node(int a, int b): place(a), dist(b){}
	bool operator<(const Node& g) const
	{
		return g.dist < dist;
	}
};

inline void dijkstra(int start)
{
	dis[start] = 0;
	priority_queue<Node> q;
	q.push(Node(start,0));
	while (!q.empty())
	{
		Node l = q.top();
		q.pop();
		vis[l.place] = true;
		for (int i = 0; i < G[l.place].size(); i++)
		{
			int nx = G[l.place][i];
			if (dis[nx] > dis[l.place] + 1)
			{
				dis[nx] = dis[l.place] + 1;
				pre[nx] = l.place;
				if (!vis[nx])
				{
					q.push(Node(nx, dis[nx]));
				}
			}
		}
	}
}

signed main()
{
	scanf("%lld %lld %lld", &n, &m, &k);
	now = m;
	for (int i = 1; i < n; i++)
	{
		int u, v;
		scanf("%lld %lld", &u, &v);
		G[u].push_back(v);
		G[v].push_back(u);
	}
	while (k--)
	{
		scanf("%lld %lld", &d, &t);
		memset(dis, 0x3f, sizeof(dis));
		memset(vis, false, sizeof(vis));
		dijkstra(now);
		if (dis[d] <= t) now = d;
		else
		{
			int pr = pre[d];
			while (dis[pr] > t) pr = pre[pr];
			now = pr;
		}
		printf("%lld ", now);
	}
	puts("");
	return 0;
}

还写了个广搜，比 Dijkstra 快一点，但还是 $48$ 分：

#include <iostream>
#include <cstdio>
#include <queue>
#include <vector>
#include <cstring>
using namespace std;

#define int long long

const int N = 1e6 + 5;

vector<int> G[N];
int n, m, k, now;
bool vis[N];
int d, t;

struct Node
{
	int place, dist;
	vector<int> path;
	Node(int a, int b, vector<int> p): place(a), dist(b), path(p){}
	Node()
	{
		place = dist = 0;
		path.clear();
	}
};

inline Node bfs(int start, int end)
{
	queue<Node> q;
	Node b;
	b.place = start, b.dist = 0;
	b.path.push_back(start);
	q.push(b);
	vis[start] = true;
	while (!q.empty())
	{
		Node l = q.front();
		q.pop();
		vector<int> copy = l.path;
		copy.push_back(l.place);
		if (l.place == end) return Node(end, l.dist, copy);
		for (register int i = 0; i < G[l.place].size(); i++)
		{
			int nx = G[l.place][i];
			if (!vis[nx])
			{
				vis[nx] = true;
				q.push(Node(nx, l.dist + 1, copy));
			}
		}
	}
}

signed main()
{
	scanf("%lld %lld %lld", &n, &m, &k);
	now = m;
	for (register int i = 1; i < n; i++)
	{
		int u, v;
		scanf("%lld %lld", &u, &v);
		G[u].push_back(v);
		G[v].push_back(u);
	}
	while (k--)
	{
		scanf("%lld %lld", &d, &t);
		memset(vis, false, sizeof(vis));
		Node g = bfs(now, d);
		if (g.dist <= t) now = d;
		else
		{
			vector<int> temp = g.path;
			int nowh = temp.size() - 1;
			while (g.dist - (temp.size() - 1 - nowh) > t) nowh--;
			now = temp[nowh];
		}
		printf("%lld ", now);
	}
	puts("");
	return 0;
}