pku 1523 SPF tarjan算法求割点

http://poj.org/problem?id=1523

求一个无向图的割点，以及输出删除该割点后形成的连通块数（删掉一个割点后会形成两个连通分块）

今天脑子被驴了。晕晕的。。。这样一个模板题整了一上午，各种粗心错误，敲了好几遍，现在看到tarjan就恶心。。。

View Code

#include <cstdio>
#include <cstring>
#include <iostream>
#define maxn 1007
using namespace std;
struct node
{
    int v;
    int next;
}g[maxn*4];
int low[maxn],dfn[maxn],cut[maxn],head[maxn];
int t,index;
int Min = 9999999;
int Max = -9999999;
int root,rtson;
void init()
{
    memset(g,0,sizeof(g));
    for (int i = 0; i < maxn; ++i)
    {
        head[i] = cut[i] = low[i] = dfn[i] = 0;
    }
    t = 1;
    index = 0;
}
void add(int u,int v)
{
    g[t].v = v;
    g[t].next = head[u];
    head[u] = t++;
}
void tarjan(int i)
{
    int j,k;
    low[i] = dfn[i] = ++index;
    for (k = head[i]; k; k = g[k].next)
    {
        j = g[k].v;
        if(!dfn[j])
        {
            tarjan(j);
            if (i == root) rtson++;
            else
            {
                if (low[i] > low[j]) low[i] = low[j];
                if (low[j] >= dfn[i]) cut[i]++;//这里来记录割点及连通块数
            }
        }
        else if(low[i] > dfn[j])
        {
            low[i] = dfn[j];
        }
    }
}
int main()
{
    //freopen("d.txt","r",stdin);
    int x,y,i,cas = 1;
    while (cin>>x)
    {
        if (!x) break;
        cin>>y;
        init();
        add(x,y); add(y,x);
        Min = min(Min,min(x,y));
        Max = max(Max,max(x,y));
        while (cin>>x)
        {
            if (!x) break;
            cin>>y;
            add(x,y); add(y,x);
            Min = min(Min,min(x,y));
            Max = max(Max,max(x,y));
        }
        root = Min;
        rtson = 0;
        tarjan(root);
        cut[root] = rtson - 1;
        bool flag = false;
        printf("Network #%d\n",cas++);
        for (i = Min; i <= Max; ++i)
        {
            if (cut[i] > 0)
            {
                flag = true;
                printf("  SPF node %d leaves %d subnets\n",i,cut[i] + 1);
            }
        }
        if (!flag)
        printf("  No SPF nodes\n");
        printf("\n");
    }
    return 0;
}