Title
Portal POJ 2186 Popular Cows
answer
T a r j a n Tarjan T a r j a n SolveSCC SCCS C C ,DAG DAGobtained after considering the shrinking pointD A G . Nodes that meet the conditions must be inDAG DAGThe only degree of D A G is0 00 ofSCC SCCS C C , theSCC SCCof the smallest topological order from the bottom upIn S C C , construct the inverse graph to judge thisSCC SCCWhether a node in S C C can traverse all points is enough. This judgment method can be used only once from any starting pointTarjan TarjanT a r j a n . Total time complexityO (N + M) O(N+M)O ( N+M)。
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 10005, maxm = 50005;
int N, M, num, dfn[maxn], low[maxn];
int top, st[maxn], scc, sc[maxn];
int tot, head[maxn], to[maxm], nxt[maxm];
int tot_r, hr[maxn], tr[maxm], nr[maxm];
bool in[maxn], vs[maxn];
inline void add(int x, int y)
{
to[++tot] = y, nxt[tot] = head[x], head[x] = tot;
tr[++tot_r] = x, nr[tot_r] = hr[y], hr[y] = tot_r;
}
void tarjan(int x)
{
low[x] = dfn[x] = ++num;
st[++top] = x, in[x] = 1;
for (int i = head[x]; i; i = nxt[i])
{
int y = to[i];
if (!dfn[y])
tarjan(y), low[x] = min(low[x], low[y]);
else if (in[y])
low[x] = min(low[x], dfn[y]);
}
if (low[x] == dfn[x])
{
++scc;
int y;
do
{
y = st[top--], sc[y] = scc;
} while (x != y);
}
}
void dfs(int x)
{
vs[x] = 1;
for (int i = hr[x]; i; i = nr[i])
{
int y = tr[i];
if (!vs[y])
dfs(y);
}
}
int main()
{
scanf("%d%d", &N, &M);
for (int i = 1, x, y; i <= M; ++i)
scanf("%d%d", &x, &y), add(x, y);
tarjan(1);
int sum = 0, x = 0;
for (int i = 1; i <= N; ++i)
if (sc[i] == 1)
++sum, x = i;
if (x)
dfs(x);
for (int i = 1; i <= N; ++i)
if (!vs[i])
{
sum = 0;
break;
}
printf("%d\n", sum);
return 0;
}
If judge SCC SCCS C C DAG DAGafter shrinking pointDoes D A G have only one out degree of1 11 node, need to construct allSCC SCCSCC。
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 10005, maxm = 50005;
int N, M, num, dfn[maxn], low[maxn];
int top, st[maxn], scc, sc[maxn], deg[maxn];
int tot, head[maxn], to[maxm], nxt[maxm];
bool in[maxn], vs[maxn];
inline void add(int x, int y)
{
to[++tot] = y, nxt[tot] = head[x], head[x] = tot;
}
void tarjan(int x)
{
low[x] = dfn[x] = ++num;
st[++top] = x, in[x] = 1;
for (int i = head[x]; i; i = nxt[i])
{
int y = to[i];
if (!dfn[y])
tarjan(y), low[x] = min(low[x], low[y]);
else if (in[y])
low[x] = min(low[x], dfn[y]);
}
if (low[x] == dfn[x])
{
++scc;
int y;
do
{
y = st[top--], sc[y] = scc;
} while (x != y);
}
}
int main()
{
scanf("%d%d", &N, &M);
for (int i = 1, x, y; i <= M; ++i)
scanf("%d%d", &x, &y), add(x, y);
for (int i = 1; i <= N; ++i)
if (!dfn[i])
tarjan(i);
int sum = 0;
for (int i = 1; i <= N; ++i)
if (sc[i] == 1)
++sum;
for (int i = 1; i <= N; ++i)
for (int j = head[i]; j; j = nxt[j])
{
int x = sc[i], y = sc[to[j]];
if (x != y)
++deg[x];
}
int cnt = 0;
for (int i = 1; i <= scc; ++i)
cnt += !deg[i];
printf("%d\n", cnt == 1 ? sum : 0);
return 0;
}