Rqnoj 272 Messenger _rqnoj

Topic description

The group of students from various schools, in order to increase friendship, the party has a word game, if a know B, then a received a message, it will pass this message to B, and all the people who know.
If a knows b,b not necessarily know a.
All the people from 1 to n numbers, give all "understanding" relationship, ask if I release a new message, then will pass a number of messages, this message back to I,1<=i<=n.
Input format

The first line in the input is two numbers n (n<1000) and M (m<10000), and a space between two numbers indicates the number of people and the number of cognitive relationships.
The next m line, two numbers a and b per line, represents a knowing B. 1<=a, B<=n. The cognitive relationship may be repeated, but a line of two numbers will not be the same.
Output format

A total of n rows in the output, one character T or F per line. Line I if it is T, I send a new message back to I; if it is F, I send a new message that does not pass back to I. Sample Input 4 6
1 2
2 3
4 1
3 1
1 3
2 3
Sample output

T
T
T
F

It's kind of like a ring-awarding problem.

But the data is small.

Direct critical table + DFS starting at each point

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int len=1111;
const int lim=11111;
int First[lim],next[lim];

int m,n;
struct Self{int x,y;} S[lim];

BOOL Flag[len];

BOOL Dfs (int i,int root)
{
    int A;
    BOOL P=false;
    Flag[i]=1;
    for (int e=first[i];e!=-1;e=next[e])
    {
        if (s[e].y==root) return true;
        if (!flag[s[e].y])
        {
            P=dfs (s[e].y,root);
            if (p) return p;
        }
    }
    return p;
}
    
int a,b;
int main ()
{
    memset (first,-1,sizeof (a));
    memset (Next,-1,sizeof (next));
    scanf ("%d%d", &m,&n);
    for (a=1;a<=n;a++)
    {
        scanf ("%d%d", &s[a].x,&s[a].y);
        Next[a]=first[s[a].x];
        first[s[a].x]=a;
    }
    
    for (a=1;a<=m;a++)
    {
        memset (flag,0,sizeof (flag));
        if (Dfs (a,a)) cout<< "T" << ' \ n ';
        else cout<< ' F ' << ' \ n ';
    }
    return 0;
}