POJ 2553--The Bottom of a Graph【scc缩点构图 && 求出度为0的scc && 输出scc中的点】

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.
Then G=(V,E) is called a directed graph.

Let n be a positive integer, and let p=(e₁,...,e_n) be a sequence of length n of edges e_i∈E such that e_i=(v_i,v_i+1) for a sequence of vertices (v₁,...,v_n+1).
Then p is called a path from vertex v₁ to vertex v_n+1 in G and we say that v_n+1 is reachable from v₁, writing (v₁→v_n+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
all nodes that are sinks, i.e., bottom(G)={v∈V|?w∈V:(v→w)?(w→v)}. You have to calculate the bottom of certain graphs.

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
numbers in the set V={1,...,v}. You may assume that 1<=v<=5000. That is followed by a non-negative integer e and, thereafter, e pairs of vertex identifiers v₁,w₁,...,v_e,w_e with
the meaning that (v_i,w_i)∈E. There are no edges other than specified by these pairs. The last test case is followed by a zero.

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

解析偷偷粘的阿宇的博客，

定义：点v是汇点须满足 --- 对图中任意点u，若v可以到达u则必有u到v的路径；若v不可以到达u，则u到v的路径可有可无。

题意：在n个点m条边的有向图里面，问有多少个点是汇点。

分析：首先若SCC里面有一个点不是汇点，那么它们全不是汇点，反之也如此。这也就意味着一个SCC里面的点要么全是，要么全不是。在求出SCC并缩点后，任一个编号为A的SCC若存在指向编号为B的SCC的边，那么它里面所有点必不是汇点（因为编号为B的SCC不可能存在指向编号为A的SCC的边）。若编号为A的SCC没有到达其他SCC的路径，那么该SCC里面所有点必是汇点。因此判断的关键在于SCC的出度是否为0.

思路：先用tarjan求出所有SCC，然后缩点后找出所有出度为0的SCC，并用数字存储点，升序排列后输出。

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#define maxn 5000
#define maxm 5000 * 5000
using namespace std;
int n, m;
struct node{
    int u, v, next;
};

node edge[maxm];

int head[maxn], cnt;
int low[maxn], dfn[maxn];
int dfs_clock;
int Stack[maxn], top;
bool Instack[maxn];
int Belong[maxn];
int scc_clock;
int in[maxn], out[maxn];
vector<int>scc[maxn];
int num[maxn];

void init(){
    cnt = 0;
    memset(head, -1, sizeof(head));
}

void addedge(int u, int v){
    edge[cnt] = {u, v, head[u]};
    head[u] = cnt++;
}

void getmap(){
    while(m--){
        int a, b;
        scanf("%d%d", &a, &b);
        addedge(a, b);
    }
}

void Tarjan(int u, int per){
    int v;
    low[u] = dfn[u] = ++dfs_clock;
    Stack[top++] = u;
    Instack[u] = true;
    for(int i = head[u]; i != -1; i = edge[i].next){
        int v = edge[i].v;
        if(!dfn[v]){
            Tarjan(v, u);
            low[u] = min(low[u], low[v]);
        }
        else if(Instack[v])
            low[u] = min(low[u], dfn[v]);
    }
    if(dfn[u] == low[u]){
        scc_clock++;
        scc[scc_clock].clear();
        do{
            v = Stack[--top];
            Instack[v] = false;
            Belong[v] = scc_clock;
            scc[scc_clock].push_back(v);
        }
        while( v != u);
    }
}

void suodian(){
    for(int i = 1; i <= scc_clock; ++i){
        out[i] = 0;
    }
    for(int i = 0; i < cnt; ++i){
        int u = Belong[edge[i].u];
        int v = Belong[edge[i].v];
        if(u != v)
            out[u]++;
    }
}

void find(){
    memset(low, 0, sizeof(low));
    memset(dfn, 0, sizeof(dfn));
    memset(Belong, 0, sizeof(Belong));
    memset(Stack, 0, sizeof(Stack));
    memset(Instack, false, sizeof(false));
    dfs_clock = scc_clock = top = 0;
    for(int i = 1; i <= n ; ++i){
        if(!dfn[i])
            Tarjan(i, i);
    }
}

void solve(){
    int k = 0;
    for(int i = 1; i <= scc_clock; ++i){
        if(out[i] == 0){
            for(int j = 0; j < scc[i].size(); ++j)
                num[k++] = scc[i][j];
        }
    }
    sort(num, num + k);
    for(int i = 0; i < k; ++i){
        if(!i)
            printf("%d", num[i]);
        else
            printf(" %d", num[i]);
    }
    printf("\n");
}

int main (){
    while(scanf("%d", &n), n){
        scanf("%d", &m);
        init();
        getmap();
        find();
        suodian();
        solve();
    }
    return 0;
}

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