pku 2117 tarjan算法求割点

http://poj.org/problem?id=2117

给出一个森林，求删掉一个点后，形成的连通块的个数。和pku1523一样的意思，只不过这里是森林了。。还要加上原来的连通块数。

注意：这里的当m == 0 时要单独考虑，因为要删掉一个点，所以== n-1 而单纯的求的话是n了。。。

View Code

#include <cstdio>
#include <cstring>
#include <iostream>
#define maxn 10010
using namespace std;
struct node
{
    int v;
    int next;
}g[4*maxn];
int low[maxn],dfn[maxn],cut[maxn],head[maxn];
int t,index,n;
int root,rtson,ans;
void add(int u,int v)
{
    g[t].v = v;
    g[t].next = head[u];
    head[u] = t++;
}
void init()
{
    memset(g,0,sizeof(g));
    for (int i = 0;i < maxn; ++i)
    {
        head[i] = low[i] = dfn[i] = cut[i] = 0;
    }
    t = 1;
    index = ans = 0;
}
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]++;
            }
        }
        else
        {
            if(low[i] > dfn[j])
            low[i] = dfn[j];
        }
    }
    ans = max(ans,cut[i]);
}
void solve()
{
     int count = 0;
    for (int i = 0; i < n; ++i)
    {
        root = i;
        rtson = 0;
        if (!dfn[root])
        {
            count++;
            tarjan(root);
        }
        cut[root] = rtson - 1;
        ans = max(cut[root],ans);
    }
    printf("%d\n",ans + count);
}
int main()
{
    int m,i,x,y;
    while (cin>>n>>m)
    {
        if (!n && !m) break;
         if (m == 0)
         {
            printf("%d\n",n-1);
            continue;
         }
        init();
        for (i = 0; i < m; ++i)
        {
            cin>>x>>y;
            add(x,y); add(y,x);
        }
        solve();
    }
}