### POJ 3694 無向圖的橋

Network
 Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 10404 Accepted: 3873

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 21 22 321 21 34 41 22 12 31 421 23 40 0`

Sample Output

`Case 1:10Case 2:20`

Source

2008 Asia Hefei Regional Contest Online by USTC   題意： 動態加邊，求橋的數目，當然不能一直dfs（tarjin）。 分析： 操作是，加邊后可以通過第一次的橋數目推出來，這條路上有橋，則，可能存在多條橋，還要繼續DFS上去。
`#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;const int maxn = 100024;vector<int> g[maxn];int low[maxn],dfn[maxn];int fa[maxn];int bin[maxn];int t;int ans;void tarjin(int u,int f) {    low[u] = dfn[u] = ++t;    fa[u] = f;    for(int i = 0; i < (int)g[u].size(); i++) {        int v = g[u][i];        if(!dfn[v]) {            tarjin(v,u);            low[u] = min(low[u],low[v]);            if(low[v]>dfn[u]) {                bin[v]++;                ans++;            }        }        else if(f!=v) {            low[u] = min(low[u],dfn[v]);            if(low[v]>dfn[u]) {         //有反向邊，不是橋                bin[v]--;                ans--;            }        }    }}void LCR(int a,int b) {    if(a==b)        return;    if(dfn[a]<dfn[b]) {        if(bin[b]==1) {            bin[b] = 0;            ans--;        }        LCR(a,fa[b]);    }    else {        if(bin[a]==1) {            bin[a] = 0;            ans--;        }        LCR(fa[a],b);    }}int main(){    //freopen("in.txt","r",stdin);    int n,m;    int kase = 1;    while(scanf("%d%d",&n,&m),n) {        for(int i = 0; i <= n; i++)            g[i].clear();        t = 0;        memset(dfn,0,sizeof(dfn));        memset(low,0,sizeof(low));        for(int i=0; i < m; i++) {            int u,v;            scanf("%d%d",&u,&v);            g[u].push_back(v);            g[v].push_back(u);        }        tarjin(1,-1);    //    printf("%d\n",ans);        printf("Case %d:\n",kase++);        int q;        scanf("%d",&q);        while(q--) {            int u,v;            scanf("%d%d",&u,&v);            LCR(u,v);            printf("%d\n",ans);        }        puts("");    }    return 0;}`

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