题目链接：
http://poj.org/problem?id=2553
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The Bottom of a Graph
Description We will use the following (standard) definitions from graph theory. Let V be a nonempty and finite set, its elements being called vertices (or nodes). Let E be a subset of the Cartesian product V×V, its elements being called edges. Let n be a positive integer, and let p=(e1,...,en) be a sequence of length n of edges ei∈E such that ei=(vi,vi+1) for a sequence of vertices (v1,...,vn+1). Here are some new definitions. A node v in a graph G=(V,E) is called a sink, if for every node w in G that is reachable from v, v is also reachable from w. The bottom of a graph is the subset of Input The input contains several test cases, each of which corresponds to a directed graph G. Each test case starts with an integer number v, denoting the number of vertices of G=(V,E), where the vertices will be identified by the integer Output For each test case output the bottom of the specified graph on a single line. To this end, print the numbers of all nodes that are sinks in sorted order separated by a single space character. If the bottom is empty, print an empty line. Sample Input 3 3 1 3 2 3 3 1 2 1 1 2 0 Sample Output 1 3 2 Source |
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题目意思：
给一幅有向图，求这样的节点i的集合,使得如果i到其他任意节点j可达的话，j也有路劲可达i。
解题思路：
tarjan+缩点+统计出度为0的强连通分量，并输出其中的节点。
代码：
//#include<CSpreadSheet.h>
#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
#define Maxn 5500
int low[Maxn],dfn[Maxn],sta[Maxn],n,m;
int dep,bc,sc,in[Maxn];
bool iss[Maxn];
vector<vector<int> >myv;
int de[Maxn];
void tarjan(int cur)
{
int ne;
low[cur]=dfn[cur]=++dep;
sta[++sc]=cur;
iss[cur]=true;
for(int i=0;i<myv[cur].size();i++)
{
ne=myv[cur][i];
if(!dfn[ne])
{
tarjan(ne);
low[cur]=min(low[cur],low[ne]);
}
else if(iss[ne]&&dfn[ne]<low[cur])
low[cur]=dfn[ne];
}
if(low[cur]==dfn[cur])
{
bc++;
do
{
ne=sta[sc--];
in[ne]=bc;
iss[ne]=false;
}while(ne!=cur);
}
}
void solve()
{
sc=bc=dep=0;
memset(iss,false,sizeof(iss));
memset(dfn,0,sizeof(dfn));
for(int i=1;i<=n;i++)
if(!dfn[i])
tarjan(i);
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
while(scanf("%d",&n)&&n)
{
scanf("%d",&m);
myv.clear();
myv.resize(n+1);
for(int i=1;i<=m;i++)
{
int a,b;
scanf("%d%d",&a,&b);
myv[a].push_back(b);
}
solve();
memset(de,0,sizeof(de));
for(int i=1;i<=n;i++)
{
for(int j=0;j<myv[i].size();j++)
{
int ne=myv[i][j];
if(in[i]!=in[ne])
de[in[i]]++;
}
}
bool fi=true;
for(int i=1;i<=n;i++)
{
if(!de[in[i]])
{
if(!fi)
putchar(' ');
else
fi=false;
printf("%d",i);
}
}
printf("\n");
}
return 0;
}