HDU 2460 Network 傻逼Tarjan-优快云博客

本文介绍了一种通过Tarjan算法解决网络中桥接问题的方法。该算法能够在每次新增连接后，快速更新网络中的桥接数量。适用于需要动态调整的大型网络。

Network

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1014 Accepted Submission(s): 206

Problem 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

题意

给你一个无向图，有多次操作。每次操作加入一条边，然后询问有多少桥。

题解

就先Tarjan求出有多少桥，然后每次操作，就在搜索树上求个lca，在求的过程中遇到的桥都会失效，因为构成了环。详见代码

#pragma comment(linker, "/STACK:102400000,102400000") 
#include<iostream>
#include<vector>
#include<cstring>
#include<string>
#include<algorithm>
#include<stack>
#define MAX_N 100005
using namespace std;

vector<int> G[MAX_N];

int father[MAX_N];

int dfn[MAX_N],low[MAX_N],ind=0;
bool vis[MAX_N];

int n,m;

int sum = 0;
bool isBridge[MAX_N];

int p;

void Tarjan(int u) {
    father[u] = p;
    dfn[u] = low[u] = ++ind;
    vis[u] = 1;
    for (int i = 0; i < G[u].size(); i++) {
        int v = G[u][i];
        if (v == p)continue;
        if (!vis[v]) {
            int tmp = p;
            p = u;
            Tarjan(v);
            p = tmp;
            low[u] = min(low[u], low[v]);
            if (low[v] > dfn[u]) {
                sum++;
                isBridge[v] = 1;
            }
        }
        else
            low[u] = min(dfn[v], low[u]);
    }
}

int main() {
    int cas = 0;
    cin.sync_with_stdio(false);
    while (cin >> n >> m) {
        if (n == 0 && m == 0)break;
        memset(vis, 0, sizeof(vis));

        for (int i = 0; i <= n; i++)G[i].clear();
        sum = 0;
        memset(isBridge, 0, sizeof(isBridge));
        ind = 0;
        memset(father, 0, sizeof(father));
        for (int i = 0; i < m; i++) {
            int u, v;
            cin >> u >> v;
            G[u].push_back(v);
            G[v].push_back(u);
        }
        p = 0;
        Tarjan(1);
        int q;
        cin >> q;
        cout << "Case " << ++cas << ":" << endl;
        while (q--) {
            int u, v;
            cin >> u >> v;
            if (dfn[u] < dfn[v])swap(u, v);
            while (dfn[u] > dfn[v]) {
                if (isBridge[u])sum--;
                isBridge[u] = 0;
                u = father[u];
            }
            while (u != v) {
                if (isBridge[v])sum--;
                isBridge[v] = 0;
                v = father[v];
            }
            cout << sum << endl;
        }
        cout << endl;
    }
    return 0;
}