POJ 3694 Network 割邊+LCA


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

/* 這題的大概思路就是,先求割邊並標記,然后縮點,形成一棵樹,然后把這顆樹上各個 結點的父結點用dfs求出來,再然后就是LCA了,因為加入某條邊后, 樹內會形成一個圈,這個圈上所有的邊將不再是橋,可以發現跟LCA的關聯。 求LCA用裸的方法就行,比較直觀些,也好操作。 實際上,這道題也不一定要縮點,如果用縮點的思路來做的話,程序將十分麻煩。 可以直接根據dfn值來進行LCA。因為,我們觀察low[v] > dfn[u]這個條件, 代表的意思就是v無法通過回邊或者通過子女到達比u點更靠前的點,那么我們只需要標記v 點即可表明割邊。在進行LCA時,由於樹的組成就是原圖中的割邊,所以在原圖中, 根據這個標記來判斷是否將割邊被轉化為了普通邊。 */
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#include<cstdio>
using namespace std;
#define maxn 100005
#define maxm 555555
#define INF 1000000000
struct Edge{ int v,next; }e[maxm];
int low[maxn],dfn[maxn],dep[maxn],vis[maxn],visx;
int tot,n,m,head[maxn];
int cnt,bridge[maxn],fa[maxn];
void Init(){
    cnt=0;tot=0;visx=0;
    memset(vis,0,sizeof vis );
    memset(dfn,0,sizeof dfn );
    memset(dep,0,sizeof dep );
    memset(bridge,0,sizeof bridge );
    memset(head,-1,sizeof head );
    for(int i=1;i<=n;i++) fa[i]=i;
}
void Add_Edge(int u,int v){ e[tot].v=v;e[tot].next=head[u];head[u]=tot++; }
void Tarjan(int u){
    dfn[u] = low[u] = ++visx;
    dep[u] = dep[fa[u]] +1;
    for(int i=head[u];i!=-1;i=e[i].next){
        int v=e[i].v;
        if(!dfn[v]){
            fa[v]=u;
            Tarjan(v);
            low[u]=min(low[u],low[v]);
            if(low[v]>dfn[u]) cnt++,bridge[v]=1;//u-v鍓茶竟
        }
        else if(v!=fa[u]) low[u]=min(low[u],dfn[v]);
    }
}
void LCA(int u,int v){
    while(dep[u]>dep[v]){
        if(bridge[u]) cnt--,bridge[u]=0;
        u=fa[u];
    }
    while(dep[v]>dep[u]){
        if(bridge[v]) cnt--,bridge[v]=0;
        v=fa[v];
    }
    while(u!=v){
        if(bridge[u]) cnt--,bridge[u]=0;
        if(bridge[v]) cnt--,bridge[v]=0;
        u=fa[u];v=fa[v];
    }
}
void Ask(){
    int q,u,v;
    scanf("%d",&q);
    while(q--){
        scanf("%d%d",&u,&v);
        LCA(u,v);
        printf("%d\n",cnt);
    }
    printf("\n");
}
int main(){
    int cas=0;
    while(scanf("%d%d",&n,&m)!=EOF){
        if(n==0&&m==0) break;
        printf("Case %d:\n", ++cas);
        Init();
        for(int u,v,i=1;i<=m;i++){
            scanf("%d%d",&u,&v);
            Add_Edge(u,v);Add_Edge(v,u);
        }
        Tarjan(1);
        Ask();
    }
    return 0;
}

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