图论——强连通分量

pku2553：

题目大意：给定一个n个节点，m条边的有向图。对某个节点v，如果从它能到达的节点w，那么从w也能到达v；这样的节点称为sink。求出图中所有的sink并排序输出。

解法：很明显和强连通分量有关，缩点后形成一张DAG图，所求既是出度为0的缩点对应的原始点。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
usingnamespace std;

constint N =5010;
constint M = N*N;

int n, m, tot, depth, top, ct, stack[N], h[N], d[N], low[N], v[M], nxt[M], index[N];
bool visit[N], instack[N];

vector <int> ans;

void add(int a, int b)
{
    v[tot] = b;
    nxt[tot] = h[a];
    h[a] = tot++;
}

void dfs(int u)
{
    int p, tv, x;
    d[u] = low[u] = depth++;
    instack[u] =true;
    stack[++top] = u;
    for(p = h[u]; p !=-1; p = nxt[p])
    {
        tv = v[p];
        if(d[tv] ==-1)
        {
            dfs(tv);
            low[u] = min(low[tv], low[u]);
        }
        elseif(instack[tv])
            low[u] = min(low[u], d[tv]);
    }
    if(d[u] == low[u])
    {
        ct++;
        while(top)
        {
            x = stack[top--];
            instack[x] =false;
            index[x] = ct;
            if(x == u)  break;
        }
    }
}

int main()
{
    int i, a, b, j;
    while(scanf("%d", &n) != EOF)
    {
        if(n ==0)  break;
        memset(h, -1, sizeof(h));
        tot =0;
        scanf("%d", &m);
        for(i =0; i < m; i++)
        {
            scanf("%d%d", &a, &b);
            add(a, b);
        }
        memset(d, -1, sizeof(d));
        depth = top = ct =0;
        memset(instack, false, sizeof(instack));
        for(i =1; i <= n; i++)
            if(d[i] ==-1)
                dfs(i);
        memset(visit, true, sizeof(visit));
        for(i =1; i <= n; i++)
            for(j = h[i]; j !=-1; j = nxt[j])
            {
                if(index[i] != index[v[j]])
                {
                    visit[index[i]] =false;
                    break;
                }
            }
        ans.clear();
        for(i =1; i <= n; i++)
            if(visit[index[i]])
                ans.push_back(i);
        sort(ans.begin(), ans.end());
        if(ans.size() >0)
        {
            for(i =0; i < ans.size()-1; i++)  printf("%d ", ans[i]);
            printf("%d\n", ans[i]);
        }
        else  printf("\n");
    }
    return0;
}