谁知晓BFS在图论中如何实现长尾词的探索？

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在面向对象编程中，两个常见的概念是is-a和has-a关系。给定两个类A和B，我们说A是B的子类，即A is-a B；如果A的一个字段类型是B，则说A has-a B。例如，我们可以这样表示：类A是类B的子类，或者类A包含一个类B的实例。

Two familiar concepts in object oriented programming are the is-a and has-a relationships. Given two classes
A and B, we say that A is-a B if A is a subclass of B; we say A has-a B if one of the fields of A is of type B.
For example, we could imagine an object-oriented language (call it ICPC++) with code like that in Figure
E.1, where the class Day is-a Time, the class Appointment is both a DateBook and a Reminder, and
class Appointment has-a Day.
class Day extends Time class Appointment extends Datebook, Reminder
{ {
… private Day date;
} …
}
Figure E.1: Two ICPC++ classes.
These two relationships are transitive. For example if A is-a B and B is-a C then it follows that A is-a C.
This holds as well if we change all the is-a’s in the last sentence to has-a’s. It also works with combinations
of is-a’s and has-a’s: in the example above, Appointment has-a Time, since it has-a Day and Day is-a
Time. Similarly, if class DateBook has-a Year then Appointment has-a Year, since Appointment
is-a DateBook.
In this problem you will be given a set of is-a and has-a relationships and a set of queries of the form A
is/has-a B. You must determine if each query is true or false.
Input:
Input starts with two integers n and m, (1 ≤ n, m ≤ 10 000), where n specifies the number of given is-a
and has-a relationships and m specifies the number of queries. The next n lines each contain one given
relationship in the form c1 r c2 where c1 and c2 are single-word class names, and r is either the string “is-a”
or “has-a”. Following this are m queries, one per line, using the same format. There will be at most 500
distinct class names in the n + m lines, and all class names in the last m lines will appear at least once in the
initial n lines. All is-a and has-a relationships between the given classes can be deduced from the n given
relationships. Is-a relationships can not be circular (apart from the trivial identity “x is-a x”).
Output
For each query, display the query number (starting at one) and whether the query is true or false.
ECNA 2017 Problem E: Is-A? Has-A? Who Knowz-A? 9
Sample Input 1 Sample Output 1
5 5
Day is-a Time
Appointment is-a Datebook
Appointment is-a Reminder
Appointment has-a Day
Datebook has-a Year
Day is-a Time
Time is-a Day
Appointment has-a Time
Appointment has-a Year
Day is-a Day
Query 1: true
Query 2: false
Query 3: true
Query 4: true
Query 5: true

注意到：
is+is->is;has+has->has; 只要有一个边是has ，路径就是has

#include<iostream> #include<cstdio> #include<algorithm> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<queue> #include<string> #include<bitset> #include<ctime> #include<deque> #include<stack> #include<sstream> typedef long long ll; using namespace std; typedef unsigned long long int ull; #define maxn 60005 #define ms(x) memset(x,0,sizeof(x)) #define Inf 0x7fffffff #define inf 0x3f3f3f3f const long long int mod = 1e9 + 7; #define pi acos(-1.0) #define pii pair<int,int> #define eps 1e-7 #define pll pair<ll,ll> ll quickpow(ll a, ll b) { ll ans = 1; a = a % mod; while (b > 0) { if (b % 2)ans = ans * a; b = b / 2; a = a * a; } return ans; } int gcd(int a, int b) { return b == 0 ? a : gcd(b, a%b); } int n, m; map<string, int>mp; int vis[1000][1000][20]; int id = 1; int iseg[1000][1000]; int haseg[1000][1000]; void num(string s) { if (mp.count(s))return; else mp[s] = id++; } void bfs(int x) { queue<pii>q; vis[x][x][0] = 1; q.push({ x,0 }); while (!q.empty()) { pii tmp = q.front(); q.pop(); for (int i = 1; i < id; i++) { if (iseg[tmp.first][i] && !vis[x][i][tmp.second]) { vis[x][i][tmp.second] = 1; q.push({ i,tmp.second }); } if (haseg[tmp.first][i] && !vis[x][i][1]) { vis[x][i][1] = 1; q.push({ i,1 }); } } } } int main() { //ios::sync_with_stdio(false); cin >> n >> m; int i, j; while (n--) { string a, b, c; cin >> a >> b >> c; num(a); num(c); if (b[0] == 'i') { iseg[mp[a]][mp[c]] = 1; } else { haseg[mp[a]][mp[c]] = 1; } } for (i = 1; i < id; i++)bfs(i); for (i = 1; i <= m; i++) { string a, b, c; cin >> a >> b >> c; if (b[0] == 'i') { printf("Query %d: %s\n", i, vis[mp[a]][mp[c]][0] ? "true" : "false"); } else { printf("Query %d: %s\n", i, vis[mp[a]][mp[c]][1] ? "true" : "false"); } } }