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就判断一下每个点前面的点的个数和后面的点的个数就可以了RT:
http://blog.csdn.net/acm_zl?viewmode=contents
int main()
{
scanf("%d%d", &n, &m);
memset(g, 0, sizeof(g));
for (int i = 0; i < m; i++)
{
int a, b;
scanf("%d%d", &a, &b);
g[a][b] = true;
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
for (int k = 1; k <= n; k++)
if (g[j][i] && g[i][k])
g[j][k] = true;
int ans = 0;
for (int i = 1; i <= n; i++)
{
int temp = 0;
for (int j = 1; j <= n; j++)
temp += g[i][j] + g[j][i];//如果g[i][j],则表示有出边,如果g[j][i],则表示有入边
//cout << temp << endl;
if (temp == n - 1)
ans++;
}
printf("%d\n", ans);
return 0;
}
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