Accessibility statistics (topological sorting)

Describe

Given a direction-free graph of N-point m-bars, the number of points that can be reached from each point is counted separately. n,m≤30000.

Input format

The first line is two integers n,m, and the next M-line is two integers x, y, and x = Y, which represents a forward edge from X.

Output format

A total of n rows representing the number of points that each point can reach.

Sample input

10 103 82 32 55 95 92 33 94 82 104 9

Sample output

1633211111

Analysis: Topology sequencing, in order to count the convenience of avoiding repeated records subsequent, using Bitset to store n nodes for the accessibility of N nodes, the binary directly through or operation to scan the topological sequence from backward forward.

#define MAX 30001int head[max],nxt[max],ver[max],deg[max];int a[max];int n,m;int cnt = 0,tot=0;bitset<30001> s[    max];void Add (int x,int y) {Ver[++tot] = y;    Nxt[tot] = head[x];    HEAD[X] = tot; deg[y]++;}    void Topsort () {queue<int>q;    for (int i= 1;i<=n;i++) if (deg[i]==0) Q.push (i);        while (Q.size ()) {int x = Q.front ();        Q.pop ();        A[++CNT] = x;            for (int i=head[x];i;i=nxt[i]) {int y = ver[i];        if (--deg[y]==0) Q.push (y);        }}}void Sol () {for (int i=cnt;i;i--) {int x = a[i];        S[X][X] = 1;            for (int j=head[x];j;j=nxt[j]) {int y=ver[j];        S[x]|=s[y];    }}}int Main () {cin>>n>>m;        for (int i=0;i<m;i++) {int x, y;        scanf ("%d%d", &x,&y);    Add (x, y);    } topsort ();    Sol ();    for (int i=1;i<=n;i++) {printf ("%d\n", S[i].count ()); } return 0;} 

Accessibility statistics (topological sorting)