7-9 Is It a Valid DFS Traversal Sequence DFS序判断

Given a directed graph and its DFS traversal sequences, you should judge if given sequences are valid. For example, with respect to the graph below,

0 1 4 2 5 3 , 0 2 5 1 4 3 and 3 2 5 1 4 0 are all valid DFS traversal sequences, but 0 2 3 1 4 5 is not.
Input Specification:

Each input file contains one test case. For each case, the first line contains two number M (<= 100, the total count of vertices), and N (the total count of edges), separated by a space. Notice that vertices are indexed from 0 to M−1. The following N lines each give an edge in the format shown below:

ID1 ID2

It denotes a directed edge from the vertex indexed ID1 to the vertex indexed ID2. Then the next line contains a number K (<= 20, the count of sequences to be judged). And the following K lines each give a DFS traversal sequence consisting of M different numbers separated by a space.
Output Specification:

For each DFS traversal sequence to be judged, output Yes if it’s valid, No if not.
Sample Input:

6 9
0 1
0 2
0 3
2 1
3 2
4 2
1 4
2 5
4 5
3
0 1 4 2 5 3
0 2 5 1 4 3
0 2 3 1 4 5

Sample Output:

Yes
Yes
No

题意 : 给定一个有向图,和一些序列,问这些序列是不是图G的一个dfs遍历结果
注意 : 即使不是从入度为0的点开始dfs也是可以的,例如5 4 2 1 0 3应该是yes

当一个点准备退栈,如果这个点有子节点未被访问则这个序列不是dfs序列

#ifdef debug
#include <time.h>
#include "/home/majiao/mb.h"
#endif

#include <iostream>
#include <algorithm>
#include <vector>
#include <string.h>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <math.h>

#define MAXN ((int)1e5+7)
#define ll long long int
#define INF (0x7f7f7f7f)
#define fori(lef, rig) for(int i=lef; i<=rig; i++)
#define forj(lef, rig) for(int j=lef; j<=rig; j++)
#define fork(lef, rig) for(int k=lef; k<=rig; k++)
#define QAQ (0)

using namespace std;

#define show(x...) \ do { \ cout << "\033[31;1m " << #x << " -> "; \ err(x); \ } while (0)

void err() { cout << "\033[39;0m" << endl; }
template<typename T, typename... A>
void err(T a, A... x) { cout << a << ' '; err(x...); }

namespace FastIO {

	char print_f[105];
	void read() {}
	void print() { putchar('\n'); }

	template <typename T, typename... T2>
		inline void read(T &x, T2 &... oth) {
			x = 0;
			char ch = getchar();
			ll f = 1;
			while (!isdigit(ch)) {
				if (ch == '-') f *= -1; 
				ch = getchar();
			}
			while (isdigit(ch)) {
				x = x * 10 + ch - 48;
				ch = getchar();
			}
			x *= f;
			read(oth...);
		}
	template <typename T, typename... T2>
		inline void print(T x, T2... oth) {
			ll p3=-1;
			if(x<0) putchar('-'), x=-x;
			do{
				print_f[++p3] = x%10 + 48;
			} while(x/=10);
			while(p3>=0) putchar(print_f[p3--]);
			putchar(' ');
			print(oth...);
		}
} // namespace FastIO
using FastIO::print;
using FastIO::read;

int n, m, Q, K, vis[MAXN], a[MAXN], ok, pn, tot, ind[MAXN];
set<int> G[MAXN];

//0 2 3 1 4 5
//注意点 : 即使不是从入度为0的点开始dfs也算是一个dfs序列
//注意点 : 即使不是从入度为0的点开始dfs也算是一个dfs序列
//注意点 : 即使不是从入度为0的点开始dfs也算是一个dfs序列
//注意点 : 即使不是从入度为0的点开始dfs也算是一个dfs序列
//注意点 : 即使不是从入度为0的点开始dfs也算是一个dfs序列
//注意点 : 即使不是从入度为0的点开始dfs也算是一个dfs序列
//注意点 : 即使不是从入度为0的点开始dfs也算是一个dfs序列
//注意点 : 即使不是从入度为0的点开始dfs也算是一个dfs序列
//注意点 : 即使不是从入度为0的点开始dfs也算是一个dfs序列
//注意点 : 即使不是从入度为0的点开始dfs也算是一个dfs序列
//注意点 : 即使不是从入度为0的点开始dfs也算是一个dfs序列
//注意点 : 即使不是从入度为0的点开始dfs也算是一个dfs序列
//注意点 : 即使不是从入度为0的点开始dfs也算是一个dfs序列
//注意点 : 即使不是从入度为0的点开始dfs也算是一个dfs序列
//注意点 : 即使不是从入度为0的点开始dfs也算是一个dfs序列
void dfs(int u) {
	vis[u] = true;
	while(pn+1<=n && !vis[a[pn+1]] && G[u].find(a[pn+1])!=G[u].end()) {
		pn ++;
		dfs(a[pn]);
	}
// printf("u=%d: ", u);
// for(auto v : G[u]) printf("(v:%d vis:%d), ", v, vis[v]);
// printf("\n");
	for(auto v : G[u]) //检查是否所有u的子节点都访问完了
		if(!vis[v]) ok = false; //有未访问的子节点,输出No
}

bool check() {
	memset(vis, false, sizeof(vis));
	pn = tot = 0;
	ok = true;
	while(++pn <= n) {
		if(!vis[a[pn]]) dfs(a[pn]);
	}
	return ok;
}

int main() {
#ifdef debug
	freopen("test", "r", stdin);
	clock_t stime = clock();
#endif
	read(n, m);
	for(int i=1; i<=m; i++) {
		int u, v;
		read(u, v);
		G[u].insert(v);
		ind[v] ++;
	}
	read(m);
	while(m--) {
		for(int i=1; i<=n; i++) read(a[i]);
		bool ok = check();
		printf("%s\n", ok ? "Yes" : "No");
	}





#ifdef debug
	clock_t etime = clock();
	printf("rum time: %lf 秒\n",(double) (etime-stime)/CLOCKS_PER_SEC);
#endif 
	return 0;
}