pku 2553 The Bottom of a Graph(tarjan缩点)

http://poj.org/problem?id=2553

题意才开始理解错了一位G中的任意一点都能到达v且v也能达到该点时v才属于bottom；如果这样最后出度为0的比为一个了，而且整个图都要缩为一点了。

其实题意是说w属于G如果v->w ==>（推出）w->v则v属于bottom。先缩点，然后统计出度为0的点，然后输出所有出度为0的点包含的点。如果出度不为0，那么v->w退不出w->v

这里的DAG比存在出度为0的点。。。所以都骂出题的二逼啊。。。

View Code

#include <cstdio>
#include <cstring>
#include <vector>
#include <stack>
#include <iostream>
#define maxn 10017
using namespace std;
int low[maxn],dfn[maxn],val[maxn],out[maxn];
bool instack[maxn],hash[maxn];
int belong[maxn];
int bcnt,index;
int n,m;
vector<int>g[maxn];
stack<int>s;
void init()
{
    for (int i = 0; i < maxn; ++i)
    {
        g[i].clear();
        low[i] = dfn[i] = belong[i] = val[i] = out[i] = 0;
        instack[i] = hash[i] = false;
    }
    index = bcnt =0;
}
void tarjan(int i)
{
    int j,k;
    low[i] = dfn[i] = ++index;
    s.push(i);
    instack[i] = true;
    for (k = 0; k < g[i].size(); ++k)
    {
        j = g[i][k];
        if (!dfn[j])
        {
            tarjan(j);
            low[i] = min(low[i],low[j]);
        }
        else if (instack[j])
        {
            low[i] = min(low[i],dfn[j]);
        }
    }
    if(low[i] == dfn[i])
    {
        bcnt++;
        do
        {
            j = s.top();
            s.pop();
            instack[j] = false;
            belong[j] = bcnt;
        } while (j != i);
    }
}
void solve()
{
        int i,j;
        for (i = 1; i <= n; ++i)
        {
            if (!dfn[i]) tarjan(i);
        }
        for (i = 1; i <= n; ++i)
        {
            for (j = 0; j < g[i].size(); ++j)
            {
                int k = g[i][j];
                if (belong[i] != belong[k])
                out[belong[i]]++;
            }
        }
        int flag[maxn] = {0};//注意输出的处理
        for (i = 1; i <= bcnt; ++i)
        {
            if (!out[i]) flag[i] = 1;
        }
        int f = 0;
        for (i = 1; i <= n; ++i)
        {
            if (flag[belong[i]])
            {
                if(f == 0)
                {
                    printf("%d",i);
                    f = 1;
                }
                else
                printf(" %d",i);
            }
        }
        printf("\n");
}
int main()
{
    //freopen("d.txt","r",stdin);
    int x,y,i;
    while (cin>>n)
    {
        if (!n) break;
        cin>>m;
        init();
        for (i = 1; i <= m; ++i)
        {
            cin>>x>>y;
            g[x].push_back(y);
        }
        solve();
    }
    return 0;
}