POJ 3694 Network(Tarjan求割邊+LCA)


Network
Time Limit: 5000MS   Memory Limit: 65536K
Total Submissions: 10969   Accepted: 4096

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

Source

 

題目大意:

給出一張圖,詢問每次加邊之后圖中有多少割邊

 

首先我們來一遍tarjan

這樣實際上形成了一棵樹

對於每次詢問,我們找出它們的LCA

在往LCA走的過程中判斷是否是割邊,如果是就取消標記

LCA暴力就可以,父親節點的信息可以在tarjan的過程中得到

 

 

 

// luogu-judger-enable-o2
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stack>
//#define getchar() (S == T && (T = (S = BB) + fread(BB, 1, 1 << 15, stdin), S == T) ? EOF : *S++)
//char BB[1 << 15], *S = BB, *T = BB;
using namespace std;
const int MAXN=1e6+10;
inline int read()
{
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
struct node
{
    int u,v,nxt;
}edge[MAXN];
int head[MAXN],num=1;
inline void AddEdge(int x,int y)
{
    edge[num].u=x;
    edge[num].v=y;
    edge[num].nxt=head[x];
    head[x]=num++;
}
int N,M;

int dfn[MAXN],low[MAXN],f[MAXN],deep[MAXN],tot=0;
int bridge[MAXN],ans=0;
void pre()
{
    for(int i=1;i<=N;i++) f[i]=i;
    memset(head,-1,sizeof(head));
    num=1;
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(bridge,0,sizeof(bridge));
    tot=0;
    ans=0;
}
void tarjan(int now,int fa)
{
    dfn[now]=low[now]=++tot;
    for(int i=head[now];i!=-1;i=edge[i].nxt)
    {
        if(!dfn[edge[i].v]) 
        {
            deep[edge[i].v]=deep[now]+1;
            f[edge[i].v]=now;
            tarjan(edge[i].v,now);
            low[now]=min(low[now],low[edge[i].v]);
            if(low[edge[i].v]>dfn[now])
            {
                bridge[edge[i].v]=1;
                ans++;
            }
        }    
        else if(edge[i].v!=fa) low[now]=min(low[now],dfn[edge[i].v]); 
    }
}
int Solve(int x,int y)
{
    if(deep[x]<deep[y]) swap(x,y);
    while(deep[x]!=deep[y])
    {
        if(bridge[x]) ans--,bridge[x]=0;
        x=f[x];    
    } 
    while(x!=y)
    {
        if(bridge[x]) ans--,bridge[x]=0;
        if(bridge[y]) ans--,bridge[y]=0;
        x=f[x];y=f[y];
    }
    return ans;
}
int main()
{
    #ifdef WIN32
    freopen("a.in","r",stdin);
    #else
    #endif
    int QWQ=0;
    while(scanf("%d%d",&N,&M)!=EOF)
    {
        if(N==0&&M==0) break;
        printf("Case %d:\n",++QWQ);
        pre();
        for(int i=1;i<=M;i++)
        {
            int x=read(),y=read();
            AddEdge(x,y);
            AddEdge(y,x);
        }
        deep[1]=1;
        tarjan(1,0);
        int Q=read();
        while(Q--)
        {
            int x=read(),y=read();
            printf("%d\n",Solve(x,y));
        }
        putchar('\n');
    }
    return 0;
}

 

 

 

 

 


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