UVALive-4839 HDU-3686 Traffic Real Time Query System 题解

题目大意：

　　有一张无向连通图，问从一条边走到另一条边必定要经过的点有几个。

思路：

　　先用tarjan将双连通分量都并起来，剩下的再将割点独立出来，建成一棵树，之后记录每个点到根有几个割点，再用RMQ求LCA计算。

　　注意：数组范围。

代码：

 #include<cstdio>
 #include<vector>
 #include<iostream>
 #include<algorithm>
 using namespace std;
 const int N=,M=;
 int u[M<<],v[M<<],nex[M<<],id[M<<],hea[N<<],dfn[N<<],low[N],st[M],sub[N],edge[M],
     fa[N<<][],f[N<<],pos[N<<],dep[N<<];
 bool vis[M<<],iscut[N],treecut[N<<];
 int tim,top,tot,cnt,num;
 vector <int> belo[N];

 int read()
 {
     int x=; char ch=getchar();
     while (ch<'' || ch>'') ch=getchar();
     while (ch>='' && ch<='') x=(x<<)+(x<<)+ch-,ch=getchar();
     return x;
 }

 void add(int x,int y) { v[cnt]=y,u[cnt]=x,nex[cnt]=hea[x],vis[cnt]=,hea[x]=cnt++; }

 void tarjan(int x)
 {
     dfn[x]=low[x]=++tim;
     for (int i=hea[x];~i;i=nex[i])
         if (!vis[i])
         {
             int y=v[i]; st[++top]=i;
             vis[i]=vis[i^]=;
             if (!dfn[y])
             {
                 tarjan(y);
                 low[x]=min(low[x],low[y]);
                 if (low[y]>=dfn[x])
                 {
                     ++sub[x],++num;
                     iscut[x]=;
                     do
                     {
                         int now=st[top--];
                         belo[u[now]].push_back(num);
                         belo[v[now]].push_back(num);
                         edge[id[now]]=num;
                         y=u[now];
                     }while (y^x);
                 }
             }
             else low[x]=min(low[x],dfn[y]);
         }
 }

 void dfs(int x)
 {
     dfn[x]=++tim; fa[++tot][]=dfn[x];
     f[tim]=x; pos[x]=tot;
     for (int i=hea[x];~i;i=nex[i])
     {
         int y=v[i];
         if (!dfn[y])
         {
             dep[y]=dep[x]+treecut[x];
             dfs(y);    fa[++tot][]=dfn[x];
         }
     }
 }

 void RMQ(int n)
 {
     for (int j=;(<<j)<=n;++j)
         for (int i=;i+j-<=n;++i)
             fa[i][j]=min(fa[i][j-],fa[i+(<<j-)][j-]);
 }

 int lca(int x,int y)
 {
     if (pos[x]<pos[y]) swap(x,y);
     int k=;
     while (<<(k+)<=pos[x]-pos[y]+) ++k;
     return f[min(fa[pos[y]][k],fa[pos[x]-(<<k)+][k])];
 }

 void wk(int n)
 {
     int i,m,x,y;
     for (tim=tot=i=;i<=n;++i) dfn[i]=;
     for (i=;i<=n;++i)
         if (!dfn[i]) dep[i]=,dfs(i);
     RMQ(tot);
     for (m=read();m--;)
     {
         x=edge[read()],y=edge[read()];
         if (x< || y<) { puts(""); continue; }
         int z=lca(x,y);
         if (x==z) printf("%d\n",dep[y]-dep[x]-treecut[x]);
         else if (y==z) printf("%d\n",dep[x]-dep[y]-treecut[y]);
              else printf("%d\n",dep[x]+dep[y]-(dep[z]<<)-treecut[z]);
     }
 }

 int main()
 {
     int n,m;
     while (~scanf("%d%d",&n,&m))
     {
         int i; cnt=top=num=tim=;
         if (!(n+m)) break;
         for (i=;i<=n;++i) hea[i]=-,dfn[i]=sub[i]=iscut[i]=,belo[i].clear();
         for (i=;i<=m;++i)
         {
             int x=read(),y=read();
             id[cnt]=i,add(x,y),id[cnt]=i,add(y,x);
         }
         for (i=;i<=n;++i)
             if (!dfn[i])
             {
                 tarjan(i);
                 if (--sub[i]<=) iscut[i]=;
             }
         for (i=;i<=num;++i) treecut[i]=;
         for (i=;i<=num+n;++i) hea[i]=-;
         cnt=;
         for (i=;i<=n;++i)
             if (iscut[i])
             {
                 sort(belo[i].begin(),belo[i].end());
                 treecut[++num]=;
                 add(belo[i][],num),add(num,belo[i][]);
                 for (int j=;j<belo[i].size();++j)
                     if (belo[i][j]^belo[i][j-]) add(belo[i][j],num),add(num,belo[i][j]);
             }
         wk(num);
     }
     return ;
 }