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NC51266. Network

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.

输入描述

The input consists of multiple test cases. Each test case starts with a line containing two integers N and .
Each of the following M lines contains two integers A and B , 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 , 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 , 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.

输出描述

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.

示例1

输入:

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

输出:

Case 1:
1
0

Case 2:
2
0

原站题解

上次编辑到这里,代码来自缓存 点击恢复默认模板

C++14(g++5.4) 解法, 执行用时: 57ms, 内存消耗: 7032K, 提交时间: 2020-08-06 16:52:48

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int head[maxn], to[maxn * 4], Next[maxn * 4];
int f[maxn], ans, pre[maxn];
int dfn[maxn], low[maxn], n, m, tot, num;
bool flag[maxn * 4];
void add(int x, int y)
{
	to[++tot] = y; Next[tot] = head[x]; head[x] = tot;
}
int getf(int x)
{
	return f[x] == x ? x : f[x] = getf(f[x]);
}
bool mer(int x, int y)
{
	int t1 = getf(x);
	int t2 = getf(y);
	if (t1 != t2)	f[t2] = t1;
	else		return false;
	return true;
}
void tarjan(int x, int in_edge, int fa)
{
	dfn[x] = low[x] = ++num; pre[x] = fa;//cout<<x<<" "<<fa<<endl;
	for (int i = head[x]; i; i = Next[i]) {
		int y = to[i];
		if (!dfn[y]) {
			tarjan(y, i, x);
			low[x] = min(low[x], low[y]);
			if (low[y] > dfn[x])
				ans++;
			else
				mer(x, y);
		}
		else	if (i != (in_edge ^ 1))
			low[x] = min(low[x], dfn[y]);
	}
}
void lca(int x, int y)
{
	if (dfn[x] < dfn[y])	swap(x, y);
	while (dfn[x] > dfn[y]) {
		//cout<<x<<" "<<dfn[x]<<" "<<y<<" "<<dfn[y]<<endl;
		if (mer(pre[x], x))	ans--;
		x = getf(pre[x]);
	}
	while (x != y) {
		if (mer(pre[y], y))	ans--;
		y = getf(pre[y]);
	}
}
int main()
{
	int T = 0;
	while (cin >> n >> m) {
		if (n == 0 && m == 0)	break;
		if (T)	printf("\n");
		printf("Case %d:\n", ++T);
		tot = 1; num = 0;
		for (int i = 1; i <= n; i++)
			dfn[i] = low[i] = head[i] = 0, f[i] = i;
		for (int i = 1; i <= m; i++) {
			int x, y;
			scanf("%d%d", &x, &y);
			add(x, y); add(y, x);
		}
		ans = 0;
		for (int i = 1; i <= n; i++)
			if (!dfn[i])	tarjan(i, 0, 0);
		int q; scanf("%d", &q);
		while (q--) {
			int t1, t2; scanf("%d%d", &t1, &t2);
			lca(t1, t2);
			cout << ans << endl;
		}
	}

}

C++(clang++11) 解法, 执行用时: 83ms, 内存消耗: 12004K, 提交时间: 2020-10-26 16:45:00

#include<bits/stdc++.h>
#define eb emplace_back
using namespace std;
const int nn=101026;
bool in[nn];
int a,b,t,n,m,q,ans,d[nn],c[nn],fa[nn];
stack<int>st;
vector<int>to[nn],son[nn];
int tarjan(int u,int fa){
	int low,cnt=0;
	in[u]=true,st.push(u);
	low=d[u]=++d[0];
	for(int i=0,v;i<(int)to[u].size();i++)
		if(d[v=to[u][i]]){
			if((v==fa&&cnt++)||v!=fa)low=min(low,d[v]);
		}else low=min(low,tarjan(v,u));
	in[u]=false;
	if(low==d[u]){
		c[u]=++c[0];
		for(;st.top()!=u;st.pop())
			c[st.top()]=c[0];
		st.pop();
	}return low;
}void dfs(int u){
	for(int i=0,v;i<(int)son[u].size();i++)
		if((v=son[u][i])!=fa[u])
			d[v]=d[fa[v]=u]+1,dfs(v);
}int main(){
	while(scanf("%d%d",&n,&m)!=EOF){
		if(n==0&&m==0)return 0;
		printf("Case %d:\n",++t);
		while(m--)
			scanf("%d%d",&a,&b),
			to[a].eb(b),to[b].eb(a);
		tarjan(1,0),ans=c[0]-1;
		for(int i=1;i<=n;i++)
			for(int j=0,u=c[i],v;j<(int)to[i].size();j++)
				if(u!=(v=c[to[i][j]]))
					son[u].eb(v),son[v].eb(u);
		for(d[1]=0,dfs(1),scanf("%d",&q);q--;){
			scanf("%d%d",&a,&b),
			a=c[a],b=c[b];
			if(d[a]<d[b])swap(a,b);
			for(;d[a]!=d[b];a=fa[a])
				if(!in[a])in[a]=true,ans--;
			for(;a!=b;a=fa[a],b=fa[b]){
				if(!in[a])in[a]=true,ans--;
				if(!in[b])in[b]=true,ans--;
			}printf("%d\n",ans);
		}putchar('\n');
		for(int i=1;i<=c[0];i++)son[i].clear();
		for(int i=1;i<=n;i++)to[i].clear();
		memset(in,false,sizeof in);
		c[0]=0,memset(d,0,sizeof d);
	}return 0;
}

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