D - Network 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

Case 1: 1 0Case 2: 2 0

求每次添加边后剩余的桥的数量

固执如我，偏要用判重边，每次输入在tarjan一遍，就是WR；
终于WR的没脾气了，就用了LCA，一发A。。。。

#include #include//割桥 #include #include #include #include #include #include using namespace std; const int maxn=1e5+7; #define mem(a,b) memset(a,b,sizeof(a))//low[]i及i的子孙相连的辈分最高的祖先节点所在的深度 //dfn[]i的深度 //tot树深，T_cou连通分量的个数 //in_stack[]是否在栈中 int dfn[maxn],low[maxn]; int tot,n,T_bridge; int fi[maxn],len; int fa[maxn]; bool bridge[maxn]; struct Adjcent { int U,V; int ne; }Adj[4*maxn]; /// void add(int u,int v) { Adj[len].U=u; Adj[len].V=v; Adj[len].ne=fi[u]; fi[u]=len++; Adj[len].U=v; Adj[len].V=u; Adj[len].ne=fi[v]; fi[v]=len++; } void init() { mem(low,0); mem(dfn,0); mem(fi,-1); mem(fa,0); mem(bridge,0); T_bridge=tot=len=0; } void tarjan(int x,int y) { dfn[x]=low[x]=++tot; for(int i=fi[x]; i!=-1; i=Adj[i].ne) { int v=Adj[i].V; if(!dfn[v]) { tarjan(v,x); fa[v]=x; //存下父节点 low[x]=min(low[x],low[v]); if(low[v]>dfn[x]) { T_bridge++; bridge[v]=true; //标记这条是桥 } } else if(v!=y) low[x]=min(low[x],dfn[v]); } } int LCA(int u,int v)//求u-->LCA-->V中桥的数量 { if(u==v) return 0; int ans=0; while(u!=v) { if(dfn[u]