POJ3694 Network

向右看齐 2021-12-24 16:27 198阅读 0赞

## 题目描述 ##

### 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. ###

## 输入格式 ##

### 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. ###

## 输出格式 ##

### 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. ###

--------------------

### 大致翻译：给你N个点M条边的无向图，并且有Q次加边，问每次加边之后图中的桥的数量。 ###

#### 显然，如果加入的边的两个端点在同一个边双内，那么桥的数量不变。所以我们先用Tarjan对原图进行边双连通分量缩点得到一棵树。 ####

#### 接着，对于两个端点不在一个边双的情况，显然桥的数量减少量等于两个端点的树上距离。我们求出树上距离，然后把两端点之间的边标记起来，即边长由原来的1改成0。每次求树上距离时就先一个个地往上爬，顺便还可以标记边。时间复杂度为O(M+QN)，可以通过本题，但显然不优。 ####

#### 既然边长变成0了，我们以后都没必要再管这些边了，所以我们可以用缩树的办法，用并查集把两个端点之间的点合并到一个集合中去，然后下次爬到这两个端点处时直接跳到LCA的位置就好了。 ####

### 时间复杂度为O(M+Qα(N))，接近线性 ###

#include<iostream> #include<cstring> #include<cstdio> #include<vector> #define maxn 100010 #define maxm 200010 using namespace std; struct edge{ int to,next; edge(){} edge(const int &_to,const int &_next){ to=_to,next=_next; } }e[maxm<<1]; int head[maxn],k; int dep[maxn],pre[maxn],fa[maxn]; int dfn[maxn],low[maxn],tot,stack[maxn],top; int uu[maxm],vv[maxm]; int n,m,t,ans; inline void init(){ for(register int i=0;i<=n;i++) head[i]=-1,dep[i]=dfn[i]=0; k=top=ans=tot=0; } inline void add(const int &u,const int &v){ e[k]=edge(v,head[u]),head[u]=k++; } void tarjan_dcc(int u,int pre){ if(dfn[u]) return; dfn[u]=low[u]=++tot,stack[++top]=u; bool flag=true; for(register int i=head[u];~i;i=e[i].next){ int v=e[i].to; if(v==pre&&flag) { flag=false; continue; } tarjan_dcc(v,u),low[u]=min(low[u],low[v]); } if(dfn[u]==low[u]){ while(stack[top]!=u) low[stack[top--]]=low[u]; top--; } } inline void rebuild(){ for(register int i=0;i<=n;i++) head[i]=-1; k=0; for(register int i=0;i<m;i++){ int u=low[uu[i]],v=low[vv[i]]; if(u!=v) add(u,v),add(v,u); } } void dfs(int u,int father,int deep){ dep[u]=deep,pre[u]=u; for(register int i=head[u];~i;i=e[i].next){ int v=e[i].to; if(v!=father) dfs(v,u,deep+1),fa[v]=u,ans++; } } int get(int x){ if(x!=pre[x]) pre[x]=get(pre[x]); return pre[x]; } inline void getlca(int u,int v){ u=get(u),v=get(v),top=0; while(u!=v){ if(dep[u]>dep[v]) stack[++top]=u,u=get(fa[u]); else stack[++top]=v,v=get(fa[v]); ans--; } while(top) pre[stack[top--]]=u; } int main(){ int T=0; while(~scanf("%d %d",&n,&m),n+m){ init(); for(register int i=0;i<m;i++){ scanf("%d %d",&uu[i],&vv[i]); add(uu[i],vv[i]),add(vv[i],uu[i]); } tarjan_dcc(1,-1),rebuild(); dfs(low[1],-1,1); scanf("%d",&t); printf("Case %d:\n",++T); while(t--){ int u,v; scanf("%d %d",&u,&v); getlca(low[u],low[v]); printf("%d\n",ans); } printf("\n"); } return 0; }

转载于:https://www.cnblogs.com/akura/p/10976147.html