POJ1236或洛谷2746或洛谷2812 Network of Schools
POJ原题链接
洛谷2746原题链接
洛谷2812(加强版)原题链接
显然在强连通分量里的所有学校都能通过网络得到软件,所以我们可以用\(tarjan\)求出强连通分量并缩点,统计缩点后每个点的入度和出度。
对于第一问,因为所有零入度的点无法通过网络得到软件,所以答案就是零入度的点的数量。
对于第二问,若有\(x\)个零入度的点,\(y\)个零出度的点,则答案就是\(\max\{x,y\}\)。注意当图被缩成一个点时,答案为\(0\)。
#include<cstdio>
using namespace std;
const int N = 110;//对于加强版,只需开大空间即可
const int M = 1e4 + 10;
int fi[N], di[M], ne[M], dfn[N], low[N], st[N], bl[N], ru[N], ch[N], l, ti, tp, SCC;
bool v[N];
inline int re()
{
int x = 0;
char c = getchar();
bool p = 0;
for (; c < '0' || c > '9'; c = getchar())
p |= c == '-';
for (; c >= '0' && c <= '9'; c = getchar())
x = x * 10 + c - '0';
return p ? -x : x;
}
inline void add(int x, int y)
{
di[++l] = y;
ne[l] = fi[x];
fi[x] = l;
}
inline int minn(int x, int y)
{
return x < y ? x : y;
}
inline int maxn(int x, int y)
{
return x > y ? x : y;
}
void tarjan(int x)
{
int i, y;
dfn[x] = low[x] = ++ti;
st[++tp] = x;
v[x] = 1;
for (i = fi[x]; i; i = ne[i])
{
y = di[i];
if (!dfn[y])
{
tarjan(y);
low[x] = minn(low[x], low[y]);
}
else
if (v[y])
low[x] = minn(low[x], dfn[y]);
}
if (!(dfn[x] ^ low[x]))
{
++SCC;
do
{
y = st[tp--];
v[y] = 0;
bl[y] = SCC;
} while (x ^ y);
}
}
int main()
{
int i, x, y, z, n, s_1 = 0, s_2 = 0;
n = re();
for (i = 1; i <= n; i++)
for (x = re(); x; x = re())
add(i, x);
for (i = 1; i <= n; i++)
if (!dfn[i])
tarjan(i);
for (z = 1; z <= n; z++)
for (i = fi[z]; i; i = ne[i])
{
y = bl[di[i]];
x = bl[z];
if (x ^ y)
{
ru[y]++;
ch[x]++;
}
}
for (i = 1; i <= SCC; i++)
{
if (!ru[i])
s_1++;
if (!ch[i])
s_2++;
}
printf("%d\n%d", s_1, SCC == 1 ? 0 : maxn(s_1, s_2));
return 0;
}
posted on 2018-09-11 20:07 Iowa_Battleship 阅读(98) 评论(0) 编辑 收藏 举报