Network
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 9434 | Accepted: 3511 |
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
2008 Asia Hefei Regional Contest Online by USTC、
原题大意：输入点的个数N和边的个数M，接下来输入整数Q代表询问的次数，之后Q行有两个整数A,B，表示将A与B相连。问每次询问后割边（桥）的数量。
解题思路1：用tarjian找割边并标记，之后用lca遍历A与B到最近公共祖先的路径，这段路上是割边的处理掉，总数-1就可以了。
表示用链表写的，各种慢，但是POJ数据太弱了，怎么写都能过。
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct list
{
int v;
list *next;
};
list *head[110010],*rear[110010];
int father[110010],times,dfn[110010],low[110010],bridge[110010],bridgenum,n;
int dep[110010];
void init()
{
int i;
memset(head,0,sizeof(head));
memset(dep,0,sizeof(dep));
memset(rear,0,sizeof(rear));
for(i=1;i<=n;++i) father[i]=i;
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(bridge,0,sizeof(bridge));
times=bridgenum=0;
}
void insert(int a,int b)
{
if(rear[a]!=NULL)
{
rear[a]->next=new list;
rear[a]=rear[a]->next;
} else head[a]=rear[a]=new list;
rear[a]->v=b;
rear[a]->next=NULL;
return;
}
void tarjian(int v)
{
bool flag=true;
dfn[v]=low[v]=++times;
dep[v]=dep[father[v]]+1;
for(list *p=head[v];p!=NULL;p=p->next)
{
if(p->v==father[v]&&flag)
{
flag=false;
continue;
}
if(!dfn[p->v])
{
father[p->v]=v;
tarjian(p->v);
low[v]=min(low[v],low[p->v]);
if(low[p->v]>dfn[v])
{
++bridgenum;
bridge[p->v]=1;
}
}else low[v]=min(low[v],dfn[p->v]);
}
}
void lca(int a,int b)
{
if(dep[a]<dep[b]) swap(a,b);
while(dep[a]!=dep[b])
{
if(bridge[a])
{
bridge[a]=0;
--bridgenum;
}
a=father[a];
}
while(a!=b)
{
if(bridge[a])
{
bridge[a]=0;
--bridgenum;
}
if(bridge[b])
{
bridge[b]=0;
--bridgenum;
}
a=father[a];
b=father[b];
}
return;
}
int main()
{
int m,i,a,b,q,ccase=0;
while(~scanf("%d%d",&n,&m),n&&m)
{
init();
for(i=0;i<m;++i)
{
scanf("%d%d",&a,&b);
insert(a,b);insert(b,a);
}
tarjian(1);
scanf("%d",&q);
printf("Case %d:\n",++ccase);
for(i=1;i<=q;++i)
{
scanf("%d%d",&a,&b);
lca(a,b);
printf("%d\n",bridgenum);
}
}
return 0;
}
解题思路2：在原来的基础上用并查集优化，这种做法是看了大神们的解题思路写出的。
将双连通的两个点弄成一个集合，这样在LCA时只要判断是否是一个集合，将桥的数量减去即可。
#include<stdio.h>
#include<string.h>
struct list
{
int v;
list *next;
};
list *head[111010],*rear[111010];
int n,m,father[111010],dep[110010],low[110010],fath[110010],bridgenum;
void init()
{
int i;
memset(head,0,sizeof(head));
memset(rear,0,sizeof(rear));
memset(dep,0,sizeof(dep));
memset(low,0,sizeof(low));
memset(fath,0,sizeof(fath));
for(i=1;i<=n;++i) father[i]=i;
bridgenum=0;
}
void insert(int a,int b)
{
if(rear[a]!=NULL)
{
rear[a]->next=new list;
rear[a]=rear[a]->next;
} else head[a]=rear[a]=new list;
rear[a]->v=b;
rear[a]->next=NULL;
}
int find(int x)
{
return father[x]==x?x:father[x]=find(father[x]);
}
void merge(int x,int y)
{
int fx=find(x);
int fy=find(y);
if(fx!=fy) father[fx]=fy;
}
void tarjian(int v,int deps)
{
bool flag=true;
dep[v]=low[v]=deps;
for(list *p=head[v];p!=NULL;p=p->next)
{
if(p->v==fath[v]&&flag)
{
flag=false;
continue;
}
if(!dep[p->v])
{
fath[p->v]=v;
tarjian(p->v,deps+1);
if(low[v]>low[p->v]) low[v]=low[p->v];
if(low[p->v]<=dep[v]) merge(p->v,v);
else bridgenum++;
}
else if(low[v]>dep[p->v]) low[v]=dep[p->v];
}
}
void judge(int v)
{
int x=find(v);
int y=find(fath[v]);
if(x!=y)
{
--bridgenum;
father[x]=y;
}
}
void lca(int u,int v)
{
while(dep[u]>dep[v])
{
judge(u);
u=fath[u];
}
while(dep[u]<dep[v])
{
judge(v);
v=fath[v];
}
while(u!=v)
{
judge(u);judge(v);
u=fath[u];v=fath[v];
}
}
int main()
{
int a,b,q,num=0;
while(~scanf("%d%d",&n,&m))
{
if(n==0&&m==0) break;
init();
while(m--)
{
scanf("%d%d",&a,&b);
insert(a,b);insert(b,a);
}
printf("Case %d:\n",++num);
tarjian(1,1);
scanf("%d",&q);
while(q--)
{
scanf("%d%d",&a,&b);
if(find(a)!=find(b)) lca(a,b);
printf("%d\n",bridgenum);
}
printf("\n");
}
return 0;
}