根据题意求的是出度为0的强连通分量的点的个数,这与POJ 2186有非常相似的地方,因为入度不方便统计,所以直接统计出度,出度为0即为正确答案。
思路:
利用Tarjan求强连通分量,并求出入度为0的强连通分量。
另外,题目要求输出从小到大,而我们知道Tarjan求强连通分量的顺序就是从小到大,所以不需要记录、排序然后输出。
CODE:
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
using namespace std;
#define MAXN 10010
#define MAXM 100010
struct Edge
{
int v, next;
}edge[MAXM];
int first[MAXN], stack[MAXN], ins[MAXN], dfn[MAXN], low[MAXN];
int belong[MAXM];
int outd[MAXN];
int n, m;
int cnt;
int scnt, top, tot;
void init()
{
cnt = 0;
scnt = top = tot = 0;
memset(first, -1, sizeof(first));
memset(ins, 0, sizeof(ins));
memset(dfn, 0, sizeof(dfn));
}
void read_graph(int u, int v)
{
edge[cnt].v = v;
edge[cnt].next = first[u];
first[u] = cnt++;
}
void dfs(int u)
{
int t;
low[u] = dfn[u] = ++tot;
ins[u] = 1;
stack[top++] = u;
for(int e = first[u]; e != -1; e = edge[e].next)
{
int v = edge[e].v;
if(!dfn[v])
{
dfs(v);
low[u] = min(low[u], low[v]);
}
else if(ins[v])
{
low[u] = min(low[u], dfn[v]);
}
}
if(low[u] == dfn[u])
{
scnt++;
do
{
t = stack[--top];
ins[t] = 0;
belong[t] = scnt;
}while(t != u);
}
}
void Tarjan()
{
for(int v = 1; v <= n; v++) if(!dfn[v])
dfs(v);
}
void solve()
{
Tarjan();
memset(outd, 0, sizeof(outd));
for(int u = 1; u <= n; u++) //n个顶点
{
for(int e = first[u]; e != -1; e = edge[e].next)
{
int v = edge[e].v;
if(belong[u] != belong[v]) outd[belong[u]]++;
}
}
for(int i = 1; i <= n; i++) if(!outd[belong[i]])
{
printf("%d ", i);
}
printf("\n");
}
int main()
{
while(scanf("%d", &n) && n)
{
init();
scanf("%d", &m);
while(m--)
{
int u, v;
scanf("%d%d", &u, &v);
read_graph(u, v);
}
solve();
}
return 0;
}
#include <cstring>
#include <cstdio>
#include <cstdlib>
using namespace std;
#define MAXN 10010
#define MAXM 100010
struct Edge
{
int v, next;
}edge[MAXM];
int first[MAXN], stack[MAXN], ins[MAXN], dfn[MAXN], low[MAXN];
int belong[MAXM];
int outd[MAXN];
int n, m;
int cnt;
int scnt, top, tot;
void init()
{
cnt = 0;
scnt = top = tot = 0;
memset(first, -1, sizeof(first));
memset(ins, 0, sizeof(ins));
memset(dfn, 0, sizeof(dfn));
}
void read_graph(int u, int v)
{
edge[cnt].v = v;
edge[cnt].next = first[u];
first[u] = cnt++;
}
void dfs(int u)
{
int t;
low[u] = dfn[u] = ++tot;
ins[u] = 1;
stack[top++] = u;
for(int e = first[u]; e != -1; e = edge[e].next)
{
int v = edge[e].v;
if(!dfn[v])
{
dfs(v);
low[u] = min(low[u], low[v]);
}
else if(ins[v])
{
low[u] = min(low[u], dfn[v]);
}
}
if(low[u] == dfn[u])
{
scnt++;
do
{
t = stack[--top];
ins[t] = 0;
belong[t] = scnt;
}while(t != u);
}
}
void Tarjan()
{
for(int v = 1; v <= n; v++) if(!dfn[v])
dfs(v);
}
void solve()
{
Tarjan();
memset(outd, 0, sizeof(outd));
for(int u = 1; u <= n; u++) //n个顶点
{
for(int e = first[u]; e != -1; e = edge[e].next)
{
int v = edge[e].v;
if(belong[u] != belong[v]) outd[belong[u]]++;
}
}
for(int i = 1; i <= n; i++) if(!outd[belong[i]])
{
printf("%d ", i);
}
printf("\n");
}
int main()
{
while(scanf("%d", &n) && n)
{
init();
scanf("%d", &m);
while(m--)
{
int u, v;
scanf("%d%d", &u, &v);
read_graph(u, v);
}
solve();
}
return 0;
}