hdu 3836 Equivalent Sets (tarjan缩点)

http://acm.hdu.edu.cn/showproblem.php?pid=3836

缩点后求每个点的入度与出度，最后结果是max（入度为0的点的个数，出度为0的点的个数）；只要整个图形成强连通图就可以证明了。。

今天真二把tarjian模板写错了，贡献了好几次wa竟然没有检查出来。。。。唉。。。

View Code

#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
#include <stack>
#define maxn 20007
using namespace std;
int low[maxn],dfn[maxn],belong[maxn],in[maxn],out[maxn];
bool instack[maxn];
int bcnt,index;
vector<int>g[maxn];
stack<int>s;
int n,m;
void init()
{
    for (int i = 0; i < maxn; ++i)
    {
        g[i].clear();
        low[i] = dfn[i] = belong[i] = out[i] = in[i] = 0;
        instack[i] = false;
    }
    bcnt = index = 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 (dfn[i] == low[i])
    {
        bcnt ++;
        do
        {
            j = s.top();
            s.pop();
            instack[j] = false;
            belong[j] = bcnt;
        }while (j != i);
    }
}
int main()
{
    int i,j,x,y,k;
    while (~scanf("%d%d",&n,&m))
    {
        init();
        if (m == 0)
        {
            printf("%d\n",n);
            continue;
        }
        for (i = 1; i <= m; ++i)
        {
            cin>>x>>y;
            g[x].push_back(y);
        }
        for (i = 1; i <= n; ++i)
        {
            if (!dfn[i]) tarjan(i);
        }
        if (bcnt == 1)
        {
            printf("0\n");
            continue;
        }
        for (i = 1; i <= n; ++i)
        {
            for (j = 0; j < g[i].size(); ++j)
            {
                k = g[i][j];
                if (belong[i] != belong[k])
                {
                    out[belong[i]]++;
                    in[belong[k]]++;
                }
            }
        }
        x = 0; y = 0;
        for (i = 1; i <= bcnt; ++i)
        {
            if (!in[i]) x++;
            if (!out[i]) y++;
        }
        printf("%d\n",max(x,y));
    }
    return 0;
}