【POJ 3694】 Network(割邊<橋>+LCA)


【POJ 3694】 Network(割邊+LCA)


Network
Time Limit: 5000MS   Memory Limit: 65536K
Total Submissions: 7971   Accepted: 2902

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≤ AB ≤ 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 ≤ ABN), 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:n個點的無向圖 初始化有m條邊

之后q次操作 每次表示在點a與點b間搭建一條邊 輸出對於q次操作 每次剩下的橋的條數


初始化能夠用tarjan算法求出橋 對於不是割邊的兩個點 就能夠算是在一個集合中 這樣用並查集就能夠進行縮點

最后生成的就是一棵樹 樹邊就是圖中的全部橋 q次詢問中 每次加邊<u,v> 假設u和v在一個集合中 說明新的邊不會造成影響

假設u和v在兩個集合中 兩個集合間的邊在加入<u,v>后就會失去橋的性質 這樣通過LCA就能夠遍歷全部兩個集合間的集合 在加上<u,v>這條邊后 這兩個集合間的集合事實上就變成了一個環 也就是能夠縮成一個點 在合並集合的過程中 就能夠把消失的橋從總和中減去了


代碼例如以下:

#include <iostream>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <list>
#include <algorithm>
#include <map>
#include <set>
#define LL long long
#define Pr pair<int,int>
#define fread() freopen("in.in","r",stdin)
#define fwrite() freopen("out.out","w",stdout)

using namespace std;
const int INF = 0x3f3f3f3f;
const int msz = 10000;
const int mod = 1e9+7;
const double eps = 1e-8;

struct Edge
{
	int v,next;
};

Edge eg[666666];
int head[233333];
int dfn[233333],low[233333];
int pre[233333];
int fa[233333];
bool vis[233333];
int tp,tim;
int ans;

void init(int n)
{
	for(int i = 1; i <= n; ++i)
		pre[i] = i;
}

int Find(int x)
{
	return pre[x] == x? pre[x]: (pre[x] = Find(pre[x]));
}

int Union(int u,int v)
{
	int k = Find(u);
	int r = Find(v);
	if(k == r) return false;
	pre[k] = r;
	return true;
}

void Tarjan(int u,int p)
{
	vis[u] = 1;
	dfn[u] = low[u] = tim++;
	int v;

	for(int i = head[u]; i != -1; i = eg[i].next)
	{
		v = eg[i].v;
		if(v == p) continue;

		if(!vis[v])
		{
			fa[v] = u;
			Tarjan(v,u);
			low[u] = min(low[u],low[v]);
			if(low[v] > dfn[u])
			{
				ans++;
			}else Union(v,u);
		}
		else low[u] = min(low[u],dfn[v]);
	}

}

void lca(int u,int v)
{
	if(dfn[v] < dfn[u]) swap(u,v);

	while(dfn[v] > dfn[u])
	{
		if(Union(v,fa[v]))
			ans--;
		v = fa[v];
	}

	while(v != u)
	{
		if(Union(u,fa[u]))
			ans--;
		u = fa[u];
	}

}

int main()
{
	//fread();
	//fwrite();

	int n,m,u,v,z = 0;

	while(~scanf("%d%d",&n,&m) && (m+n))
	{
		memset(head,-1,sizeof(head));
		tim = tp = 0;
		init(n);

		while(m--)
		{
			scanf("%d%d",&u,&v);
			eg[tp].v = v;
			eg[tp].next = head[u];
			head[u] = tp++;

			eg[tp].v = u;
			eg[tp].next = head[v];
			head[v] = tp++;
		}

		memset(vis,0,sizeof(vis));
		ans = 0;
		fa[1] = 1;
		Tarjan(1,1);

		int q;
		scanf("%d",&q);

		printf("Case %d:\n",++z);

		while(q--)
		{
			scanf("%d%d",&u,&v);
			lca(u,v);
			printf("%d\n",ans);
		}
		puts("");
	}

	return 0;
}







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