poj 3694&&hdu 2460 (手写递归栈版的tarjan)

题意：求加入一条边后剩余的桥。

题解：加入的边形成的环里的桥都不存在了。

#pragma comment(linker, "/STACK:10240000000000,10240000000000")
#include <iostream>
#include <vector>
#include <stack>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=110009;
const int M=210005;
int low[maxn];
int dfn[maxn];
int n,m;
vector<int>p[maxn];
stack<int>s;
int aa,bb;
int qu;
int vis[maxn];
int fa[maxn];
int b[maxn];
int tmp=1;
int ecnt;
struct Node
{
    int u,v,next;
} edge[maxn*10];

int head[maxn];
int cnt;
void init()
{
    tmp=1;
    for(int i=1; i<=n; i++)
     fa[i]=i;
    memset(b,0,sizeof(b));
    memset(vis,0,sizeof(vis));
    cnt=1;
    ecnt=0;
    memset(head,-1,sizeof(head));
}
void addedge(int u,int v)
{
        if(u==v)return;
        edge[ecnt].u = u;
        edge[ecnt].v = v;
        edge[ecnt].next = head[u];
        head[u] = ecnt++;
}

int tp;
int cur[maxn];
int st[maxn*10];
/*void tarjan(int u)
{
    visit[u]=1;
    dfn[u]=low[u]=tmp++;
    for(int i=0; i<p[u].size(); i++)
    {
        int v=p[u][i];
        if(!visit[v])
        {
            fa[v]=u;
            tarjan(v);
            low[u]=min(low[u],low[v]);
            if(low[v]>dfn[u])
            {
                cnt++;
                b[v]=1;
            }
        }
        else if(v != fa[u] && visit[v])
        {
            low[u]=min(low[u],dfn[v]);
        }
    }
}
*/
void tarjan()
{
    for(int i = 1; i < maxn; ++ i)
        cur[i] = head[i];
    st[++tp] = 1;
    while(tp)
    {
        int k = st[tp];
        if(!vis[k])
        {
            dfn[k] = low[k] =  tmp++;
            vis[k] = 1;
        }
        for(; cur[k]!=-1; cur[k]=edge[cur[k]].next)
        {
            int v = edge[cur[k]].v;
            if(!vis[v])
            {
                fa[v] = k;
                break;
            }
            else if(vis[v]==1&&v!=fa[k])
                 low[k]=min(dfn[v],low[k]);
        }
        if((cur[k]==-1)&&tp>1)
        {
            low[st[tp-1]] = min(low[k],low[st[tp-1]]);
            if(low[k]>dfn[st[tp-1]])
            {
                b[k] = 1;
                cnt++;
            }
        }
        if(cur[k]==-1)
        {
            vis[k] = 2;
            tp--;
        }
        else st[++tp] = edge[cur[k]].v;
    }
}

void LCA(int u,int v)
{
    while(dfn[u]>dfn[v])
    {
        if(b[u])
        {
            cnt--;
            b[u]=0;
        }
        u=fa[u];
    }
    while(dfn[v]>dfn[u])
    {
        if(b[v])
        {
            cnt--;
            b[v]=0;
        }
        v=fa[v];
    }
    while(v!=u)
    {
        if(b[u])
        {
            cnt--;
            b[u] = 0;
        }
        if(b[v])
        {
            cnt--;
            b[v] = 0;
        }
        u = fa[u];
        v = fa[v];
    }
}

int main()
{
    int cas=1;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(!n&&!m)break;
        init();
        for(int i=0; i<m; i++)
        {
            scanf("%d%d",&aa,&bb);
            addedge(aa,bb);
            addedge(bb,aa);
        }
        tarjan();
        for(int i=1;i<=n;i++)
         cout<<"i : "<<i<<" "<<dfn[i]<<endl;
        cin>>qu;
        cout<<"Case "<<cas++<<":"<<endl;
        while(qu--)
        {
            scanf("%d%d",&aa,&bb);
            LCA(aa,bb);
            printf("%d\n",cnt-1);
        }
        cout<<endl;
    }
    return 0;
}