hdu4338 Simple Path

Everybody knows that totalfrank has absolutely no sense of direction. Getting lost in the university or nearly supermarket is very common for him. We always worry about whether he can find his way back into our sweet base whenever he goes out alone for his class. In general, if totalfrank get lost again, we need to check his starting point and destination just in order to find out where he could be (you know this task is very common for us).
Unfortunately, poor totalfrank sometimes forgot taking his mobile phone, when this situation happens, we can’t get in touch with him. But it is so lucky that totalfrank can remember places where he had gone before in his trip from his starting point and destination at this trip so that he won’t go to such place again (he can’t remember places which he had gone during his previous trip). As we are all familiar with map, we can find out which place he couldn’t be.
However, totalfrank can always get lost, doing this same boring work makes us sleepy. So we ask totalfrank for all possible starting point and destination for him, and try to find out how many places he wouldn’t be when he chooses any pair of starting point and destination. Here comes the problem, since our university’s map is so complex (there can be many buildings which can be considered as points in our university, some pair of these point has a way while others hasn’t), we need a program to help us work out this problem.

 #pragma comment(linker, "/STACK:102400000,102400000")
 #include <string.h>
 #include <stdio.h>
 #include <math.h>
 #include <algorithm>
 const int MAXN=1e5+;

 struct edge{
     int u,v,n;
 }e1[MAXN*],e2[MAXN*];

 int f1[MAXN],f2[MAXN*],es1,es2;
 int n,m,q,tu,tv;

 void addedge1(int u,int v){
     e1[es1].u=u,e1[es1].v=v,e1[es1].n=f1[u],f1[u]=es1++;
 }

 void addedge2(int u,int v){
     e2[es2].u=u,e2[es2].v=v,e2[es2].n=f2[u],f2[u]=es2++;
 }

 int p[MAXN*];

 int find(int x){return x==p[x]?x:p[x]=find(p[x]);}

 void merge(int x,int y){p[find(x)]=find(y);}

 int sum[MAXN*],tsum[MAXN*],dis[MAXN];
 int lca_f[MAXN*],lca_b[MAXN*],lca_p[MAXN*],rid;
 int dminv[MAXN*][],dminid[MAXN*][];

 void dp(int u,int f,int dd,int tot){
     dis[u]=dd,tsum[u]=tot+sum[u];
     lca_f[++rid]=u,lca_b[rid]=dd,lca_p[u]=rid;
     for(int i=f2[u];i!=-;i=e2[i].n){
         int v=e2[i].v;
         if(v==f)continue;
         dp(v,u,dd+,tot+sum[u]);
         lca_f[++rid]=u,lca_b[rid]=dd;
     }
 }

 void makermq(){
     rid=;
     dp(,-,,);
     for(int i=;i<=rid;i++)dminv[i][]=lca_b[i],dminid[i][]=i;
     int maxj=(int)(log(rid+1.0)/log(2.0));
     for(int j=;j<=maxj;j++){
         int maxi=rid+-(<<j);
         for(int i=;i<=maxi;i++){
             if(dminv[i][j-]<dminv[i+(<<(j-))][j-]){
                 dminv[i][j]=dminv[i][j-];
                 dminid[i][j]=dminid[i][j-];
             }else{
                 dminv[i][j]=dminv[i+(<<(j-))][j-];
                 dminid[i][j]=dminid[i+(<<(j-))][j-];
             }
         }
     }
 }

 int lca(int x,int y){
     if(lca_p[x]>lca_p[y])std::swap(x,y);
     x=lca_p[x],y=lca_p[y];
     int k=(int)(log(y-x+1.0)/log(2.0));
     int xx=dminv[x][k]<dminv[y+-(<<k)][k]?dminid[x][k]:dminid[y+-(<<k)][k];
     return lca_f[xx];
 }

 int dfn[MAXN],low[MAXN],cid[MAXN],stk[MAXN],col[MAXN],top,ind,cls,tmp;
 int cal[MAXN*];

 void dfs_cutpnt(int u,int f,int root){
     dfn[u]=low[u]=++ind;
     int cnt=;
     int flag=;
     for(int i=f1[u];i!=-;i=e1[i].n){
         int v=e1[i].v;
         if(v==f&&!flag){flag=;continue;}
         if(!dfn[v]){
             cnt++;
             dfs_cutpnt(v,u,root);
             if(low[v]<low[u])low[u]=low[v];
             if(u==root&&cnt>&&cid[u]==)cid[u]=++cls,sum[cls]=;
             else if(u!=root&&low[v]>=dfn[u]&&cid[u]==)cid[u]=++cls,sum[cls]=;
         }else if(dfn[v]<low[u])low[u]=dfn[v];
     }
 }

 void dfs_tarjan(int u,int f){
     low[u]=dfn[u]=++ind;
     stk[++top]=u;
     int flag=;
     for(int i=f1[u];i!=-;i=e1[i].n){
         int v=e1[i].v;
         if(v==f&&!flag){flag=;continue;}
         if(!dfn[v]){
             dfs_tarjan(v,u);
             if(low[v]<low[u])low[u]=low[v];
             if(low[v]>=dfn[u]){
                 sum[++cls]=,col[u]=cls;
                 do{
                     tmp=stk[top--],col[tmp]=cls,++sum[cls];
                     if(cid[tmp]){addedge2(cid[tmp],cls);addedge2(cls,cid[tmp]);merge(cid[tmp],cls);}
                 }while(tmp!=v);
                 if(cid[u]){addedge2(cid[u],cls);addedge2(cls,cid[u]);merge(cid[u],cls);}
             }
         }else if(dfn[v]<low[u])low[u]=dfn[v];
     }
 }

 int size;

 void makegraph(){
     //找割点
     memset(dfn,,sizeof dfn);
     memset(low,,sizeof low);
     memset(cid,,sizeof cid);
     cls=ind=;
     //找双联通分量并建图
     for(int i=;i<n;i++)dfs_cutpnt(i,-,i);
     memset(dfn,,sizeof dfn);
     memset(low,,sizeof low);
     memset(col,,sizeof col);
     top=ind=;
     for(int i=;i<n;i++)dfs_tarjan(i,-);
     //将森林补成树,便于dp以及查询
     memset(cal,,sizeof cal);
     for(int i=;i<=cls;i++){
         if(cal[find(i)]==){
             cal[find(i)]=;
             addedge2(,i);
         }
     }
 }

 int main(){
     //freopen("test.in","r",stdin);
     int cas=;
     while(scanf("%d%d",&n,&m)!=EOF){
         memset(f1,-,sizeof f1);
         memset(f2,-,sizeof f2);
         for(int i=;i<=*n;i++)p[i]=i;
         es1=es2=;

         for(int i=;i<m;i++){
             scanf("%d%d",&tu,&tv);
             addedge1(tu,tv);
             addedge1(tv,tu);
         }

         //转化成双联通与割点相邻的图
         makegraph();
         //lca转化成rmq
         makermq();

         printf("Case #%d:\n",cas++);
         scanf("%d",&q);
         while(q--){
             scanf("%d%d",&tu,&tv);
             //起点和终点重合
             if(tu==tv)printf("%d\n",n-);
             else{
                 //如果是割点的话就一定要用割点对应的点,因为割点会被染成不同的颜色!
                 tu=cid[tu]?cid[tu]:col[tu];
                 tv=cid[tv]?cid[tv]:col[tv];
                 //孤立点或者不在同一个联通块中
                 if(tu==||tv==||find(tu)!=find(tv)){
                     printf("%d\n",n);
                 }else{
                     int fa=lca(tu,tv);
                     int ans=tsum[tu]+tsum[tv]-*tsum[fa]+sum[fa];
                     ans-=(dis[tu]+dis[tv]-*dis[fa]);
                     printf("%d\n",n-ans);
                 }
             }
         }
         printf("\n");
     }
     return ;
 }