PAT (Top Level) Practise 1008 Airline Routes (35)

1008. Airline Routes (35)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CHEN, Yue

Given a map of airline routes, you are supposed to check if a round trip can be planned between any pair of cities.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers N (2<= N <= 104) and M (<=6N), which are the total number of cities (hence the cities are numbered from 1 to N) and the number
of airline routes, respectively. Then M lines follow, each gives the information of a route in the format of the source city index first, and then the destination city index, separated by a space. It is guaranteed that the source is never the same as the destination.

After the map information, another positive integer K is given, which is the number of queries. Then K lines of queries follow, each contains a pair of distinct cities' indices.

Output Specification:

For each query, output in a line "Yes" if a round trip is possible, or "No" if not.
Sample Input:

12 19
3 4
1 3
12 11
5 9
6 2
3 2
10 7
9 1
7 12
2 4
9 5
2 6
12 4
11 10
4 8
8 12
11 8
12 7
1 5
20
11 4
12 7
3 6
2 3
5 3
3 9
4 3
8 3
8 10
10 11
7 8
7 1
9 5
1 9
2 6
3 1
3 12
7 3
6 9
6 8

Sample Output:

Yes
Yes
No
No
No
No
No
No
Yes
Yes
Yes
No
Yes
Yes
Yes
No
No
No
No

No
题意是给出一个有向图，然后问两个点在不在一个环上。直接上tarjan强连通即可，是一道裸题，只要判断low【x】和low【y】是不是相同即可#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<stack>
#include<algorithm>
#include<bitset>
#include<functional>
using namespace std;
typedef unsigned long long ull;
typedef long long LL;
const int maxn = 2e5 + 10;
int n, m, x, y, tot, t;
int dfn[maxn], low[maxn];
int ft[maxn], nt[maxn], v[maxn];
int vis[maxn];
stack<int> p;

void tarjan(int x)
{
dfn[x] = low[x] = ++t;
p.push(x); vis[x] = 1;
for (int i = ft[x]; i != -1; i = nt[i])
{
if (!dfn[v[i]])
{
tarjan(v[i]);
low[x] = min(low[x], low[v[i]]);
}
else if (vis[v[i]]) {
low[x] = min(low[x], dfn[v[i]]);
}
}
if (low[x] == dfn[x])
{
int u;
do
{
u = p.top(); p.pop();
low[u] = low[x];
vis[u] = 0;
} while (u != x);
}
}

int main()
{
while (scanf("%d%d", &n, &m) != EOF)
{
tot = t = 0;
while (!p.empty()) p.pop();
for (int i = 1; i <= n; i++) ft[i] = -1, dfn[i] = vis[i] = 0;
while (m--)
{
scanf("%d%d", &x, &y);
v[tot] = y; nt[tot] = ft[x]; ft[x] = tot++;
}
for (int i = 1; i <= n; i++) if (!dfn[i]) tarjan(i);
scanf("%d", &m);
while (m--)
{
scanf("%d%d", &x, &y);
printf("%s\n", low[x] == low[y] ? "Yes" : "No");
}
}
return 0;
}