『图的遍历与判环』CF864F Cities Excursions

$\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{b}\mathrm{l}\mathrm{e}\mathrm{m}$

There are $n$ cities in Berland. Some pairs of them are connected with $m$ directed roads. One can use only these roads to move from one city to another. There are no roads that connect a city to itself. For each pair of cities $(x,y)$ there is at most one road from $x$ to $y$ .

A path from city $s$ to city $t$ is a sequence of cities $p_1$ , $p_2$ , … , $p_k$ , where $p_1=s$ , $p_k=t$ , and there is a road from city $p_i$ to city $p_{i+1}$ for each $ i $ from $1$ to $k-1$ . The path can pass multiple times through each city except $t$ . It can't pass through $t$ more than once.

A path $p$ from $s$ to $t$ is ideal if it is the lexicographically minimal such path. In other words, $p$ is ideal path from $s$ to $t$ if for any other path $q$ from $s$ to $t$ $p_i$<$q_i$ , where $i$ s the minimum integer such that $p_i\ne q_i$ .

There is a tourist agency in the country that offers $q$ unusual excursions: the $j$ -th excursion starts at city $s_j$ and ends in city $t_j$ .

For each pair $s_j$ , $t_j$ help the agency to study the ideal path from $s_j$ to $t_j$ . Note that it is possible that there is no ideal path from $s_j$ to $t_j$ . This is possible due to two reasons:

• there is no path from $s_j$ to $t_j$ ;
• there are paths from $s_j$ to $t_j$ , but for every such path $p$ there is another path $q$ from $s_j$ to $t_j$ , such that $p_i$>$q_i$ , where $i$ is the minimum integer for which $p_i\ne q_i$ .

The agency would like to know for the ideal path from $s_j$ to $t_j$ the $k_j$ -th city in that path (on the way from $s_j$ to $t_j$ ).

For each triple $s_j$ , $t_j$ , $k_j$ ( $i\le j\le k$ ) find if there is an ideal path from $s_j$ to $t_j$ and print the $k_j$ -th city in that path, if there is any.

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$\mathrm{T}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{s}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{e}$

给定一个有向图,询问由$s$到$t$的字典序最小的路径的第$k$个点

若字典序最小的路径构成环输出$-1$.

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$\mathrm{S}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}$

这道题的思路还是很直白的：

• 我们将某一个点所连接点的标号排序，那么就转化为了图的遍历。
• 每次走到某一个点时，统计起点到该点的路径，那么对于询问$(s,t,k)$，只要满足统计路径上点的数量大于等于$k$，且当前点不是从上出来的，那么路径上的第$k$个点就是合法的答案。

那么接下来就是如何实现的问题反正我是写不来

• 对于某一个已经遍历过的节点，若第二题遍历到它字典序一定不是最小的，因此我们需要每一个节点都遍历一个。
• 如果遇到的某一个点处在已经统计过的节点中，那么我们是环上的哪一个根据固定的路径走，都是从环上出来的，肯定无解。因此，我们要在已经统计过的点打上标记，直到从这个点离开以后才能撤销标记。（说明已经不存在环上的路径了）
• 怎么这么妙啊啊啊啊

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$\mathrm{C}\mathrm{o}\mathrm{d}\mathrm{e}$

#include 

using namespace std;
const int N = 3010;

int n, m, Q, loop;
int tag[N], cnt[N], ans[N*N];
vector < int > a[N], Now;
vector < pair<int,int> > q[N][N];

int read(void)
{
	int s = 0; char c = getchar();
	while (c < '0' || c > '9') c = getchar();
	while (c >= '0' && c <= '9') s = s * 10 + c - 48, c = getchar();
	return s; 
}

void Dfs(int root, int x)
{
	tag[x] = x * 2 - 1;
	Now.push_back(x);
	for (int i=0;i<q[root][x].size();++i)
	{
		int k = q[root][x][i].first;
		int id = q[root][x][i].second;
		if (k <= Now.size() and loop == 0) 
			ans[id] = Now[k-1]; 
	}
	for (int i=0;i<a[x].size();++i)
	{
		int y = a[x][i];
		if (tag[y] < 2 * y - 1) Dfs(root, y);
		else if (tag[y] == 2 * y - 1) loop ++, cnt[y] ++;
	} 
	Now.pop_back();
	loop -= cnt[x];
	cnt[x] = 0;
	tag[x] = x * 2;
	return;
}

int main(void)
{
	n = read(), m = read(), Q = read();
	for (int i=1;i<=m;++i)
	{
		int x = read(), y = read();
		a[x].push_back(y);
	}
	for (int i=1;i<=n;++i)
		sort(a[i].begin(), a[i].end());
	for (int i=1;i<=Q;++i)
	{
		int x = read(), y = read(), k = read();
		q[x][y].push_back({k, i});
		ans[i] = -1;
	}
	for (int i=1;i<=n;++i)
	{
		loop = 0;
		memset(tag, 0, sizeof tag);
		Dfs(i, i);
	} 
	for (int i=1;i<=Q;++i) printf("%d ", ans[i]);
	return 0;
}