[POJ3694]Network
Description
A network administrator manages a large network. The network consists of N computers and M links between pairs of computers. Any pair of computers are connected directly or indirectly by successive links, so data can be transformed between any two computers. The administrator finds that some links are vital to the network, because failure of any one of them can cause that data can't be transformed between some computers. He call such a link a bridge. He is planning to add some new links one by one to eliminate all bridges.
You are to help the administrator by reporting the number of bridges in the network after each new link is added.
Input
The input consists of multiple test cases. Each test case starts with a line containing two integers N(1 ≤ N ≤ 100,000) and M(N - 1 ≤ M ≤ 200,000).
Each of the following M lines contains two integers A and B ( 1≤ A ≠ B ≤ N), which indicates a link between computer A and B. Computers are numbered from 1 to N. It is guaranteed that any two computers are connected in the initial network.
The next line contains a single integer Q ( 1 ≤ Q ≤ 1,000), which is the number of new links the administrator plans to add to the network one by one.
The i-th line of the following Q lines contains two integer A and B (1 ≤ A ≠ B ≤ N), which is the i-th added new link connecting computer A and B.
The last test case is followed by a line containing two zeros.
Output
For each test case, print a line containing the test case number( beginning with 1) and Q lines, the i-th of which contains a integer indicating the number of bridges in the network after the first i new links are added. Print a blank line after the output for each test case.
Sample Input
3 2
1 2
2 3
2
1 2
1 3
4 4
1 2
2 1
2 3
1 4
2
1 2
3 4
0 0
Sample Output
Case 1:
1
0
Case 2:
2
0
题意
给定一张(N)个点(M)条边的无向连通图，然后执行(Q)次操作，每次向图中添加一条边，并且询问当前无向图中“桥”的数量。
核心知识：边双联通分量
先一遍(tarjan)把原图中的桥全部标记，然后(dfs)把各个边双联通分量建成树。
考虑到每次加边：
1.两端点在同一个边双联通分量中，对答案不造成影响。
2.两端点不在同一个边双联通分量，因为在树中，所以可以形成一个环，两个边双联通分量之间的路径不再是桥，在总答案中减去。
因为本题时间比较宽裕，所以我们暴力往上跳就可以过了，但是还可以用并查集优化，但是没有打出来，下次再填坑吧。。。
边双联通分量的求法
求出无向图中所有的桥，把桥“删除”，无向图会分成若干个联通块，每一个联通块就是一个边双联通分量。求边双联通分量的时间复杂度:(O(n))。
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int read()
{
int x=0,w=1;char ch=getchar();
while(ch>'9'||ch<'0') {if(ch=='-')w=-1;ch=getchar();}
while(ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
return x*w;
}
const int N=100010;
int n,m,q,t,cnt=1,visnum,num,ans,a,b;
int head[2*N],dfn[N],low[N],fa[2*N],x[2*N],y[2*N],deep[2*N];
bool bridge[4*N],vis[N],dcc[N];
struct node{
int to,next;
}edge[8*N];
void add(int x,int y)
{
cnt++;edge[cnt].to=y;edge[cnt].next=head[x];head[x]=cnt;
}
int gfa(int x){if(fa[x]==x)return x;return fa[x]=gfa(fa[x]);}
void tarjan(int k,int last)
{
dfn[k]=low[k]=++visnum;
for(int i=head[k];i;i=edge[i].next)
{
int v=edge[i].to;
if(!dfn[v])
{
tarjan(v,i);low[k]=min(low[k],low[v]);
if(low[v]>dfn[k])
{
bridge[i]=bridge[i^1]=1;ans++;
}
}
else if(i!=(last^1))low[k]=min(low[k],dfn[v]);
}
}
void dfs1(int k)
{
vis[k]=1;fa[k]=num;
for(int i=head[k];i;i=edge[i].next)
{
int v=edge[i].to;if(vis[v]||bridge[i])continue;
dfs1(v);
}
}
void dfs2(int k,int f)
{
for(int i=head[k];i;i=edge[i].next)
{
int v=edge[i].to;if(v==f)continue;
deep[v]=deep[k]+1;fa[v]=k;dcc[v]=1;dfs2(v,k);
}
}
int jump(int xx,int yy)
{
int qwe=0;if(deep[xx]<deep[yy]) swap(xx,yy);
while(deep[xx]>deep[yy])
{
if(dcc[xx]) dcc[xx]=0,qwe++;xx=fa[xx];
}
if(xx==yy)return qwe;
while(xx!=yy)
{
if(dcc[xx]) dcc[xx]=0,qwe++;xx=fa[xx];
if(dcc[yy]) dcc[yy]=0,qwe++;yy=fa[yy];
}
return qwe;
}
int main()
{
while(1)
{
n=read();m=read();if(n==0)break;t++;num=n;
for(int i=1;i<=m;i++)
{
x[i]=read();y[i]=read();
add(x[i],y[i]);add(y[i],x[i]);
}
tarjan(1,0);
for(int i=1;i<=n;i++)
{
if(!vis[i]){num++;fa[num]=num;dfs1(i);}
}
for(int i=1;i<=m;i++)
{
int xx=gfa(x[i]),yy=gfa(y[i]);
if(xx!=yy) add(xx,yy),add(yy,xx);
}
deep[n+1]=1;dfs2(n+1,0);q=read();
printf("Case %d:
",t);
for(int j=1;j<=q;j++)
{
a=read();b=read();
int xx=fa[a],yy=fa[b];
if(xx!=yy) ans-=jump(xx,yy);printf("%d
",ans);
}
printf("
");
}
}