Description
It's almost summer time, and that means that it's almost summer construction time! This year, the good people who are in charge of the roads on the tropical island paradise of Remote Island would like to repair and upgrade the various roads that lead between the various tourist attractions on the island.
The roads themselves are also rather interesting. Due to the strange customs of the island, the roads are arranged so that they never meet at intersections, but rather pass over or under each other using bridges and tunnels. In this way, each road runs between two specific tourist attractions, so that the tourists do not become irreparably lost.
Unfortunately, given the nature of the repairs and upgrades needed on each road, when the construction company works on a particular road, it is unusable in either direction. This could cause a problem if it becomes impossible to travel between two tourist attractions, even if the construction company works on only one road at any particular time.
So, the Road Department of Remote Island has decided to call upon your consulting services to help remedy this problem. It has been decided that new roads will have to be built between the various attractions in such a way that in the final configuration, if any one road is undergoing construction, it would still be possible to travel between any two tourist attractions using the remaining roads. Your task is to find the minimum number of new roads necessary.
Input
The first line of input will consist of positive integers n and r, separated by a space, where 3 ≤ n ≤ 1000 is the number of tourist attractions on the island, and 2 ≤ r ≤ 1000 is the number of roads. The tourist attractions are conveniently labelled from 1 to n. Each of the following r lines will consist of two integers, v and w, separated by a space, indicating that a road exists between the attractions labelled v and w. Note that you may travel in either direction down each road, and any pair of tourist attractions will have at most one road directly between them. Also, you are assured that in the current configuration, it is possible to travel between any two tourist attractions.
Output
One line, consisting of an integer, which gives the minimum number of roads that we need to add.
Sample Input
Sample Input 1
10 12
1 2
1 3
1 4
2 5
2 6
5 6
3 7
3 8
7 8
4 9
4 10
9 10
Sample Input 2
3 3
1 2
2 3
1 3
Sample Output
Output for Sample Input 1
2
Output for Sample Input 2
0
Source
題目大意：給定一個(gè)雙向連通的公路網(wǎng)，當(dāng)某些公路路段檢修的時(shí)候可能會(huì)由于該段公路不通，可能會(huì)使某些旅游點(diǎn)之間無法通行，求至少新建多少條公路，使得任意對(duì)一段公路進(jìn)行檢修的時(shí)候，所有的旅游景點(diǎn)之間仍然暢通；
分析：檢修某一路段導(dǎo)致公路網(wǎng)不暢通的原因必然是該段公路在圖中是橋（割邊），因此完全暢通的方法就是，加最若干條邊，使圖中不存在橋。先找出圖中所有的雙連通分量，將雙連通分量進(jìn)行縮點(diǎn)，得到一個(gè)樹形圖，求出這個(gè)樹形圖中度為1的點(diǎn)的個(gè)數(shù)，新加邊的條數(shù)即是（度為1的點(diǎn)數(shù)目+1）/2，考慮到題目只要求求度為1的點(diǎn)數(shù)目，因此可以部分縮點(diǎn)，利用并查集，保存每個(gè)割邊的頂點(diǎn)，統(tǒng)計(jì)每個(gè)頂點(diǎn)在并查集中的代表元的度數(shù)即可。
Sample 1中存在4個(gè)雙連通分量：{1}，{2，5，6}，{3，7，8}，{4，9，10}，進(jìn)行縮點(diǎn)之后，求得一個(gè)4個(gè)節(jié)點(diǎn)的樹形圖，其中一個(gè)點(diǎn)的度數(shù)為3，其余3個(gè)點(diǎn)的度數(shù)為1，得到需要加的邊的數(shù)目為(3+1)/2=2。
#include <iostream>
#include <vector>
using namespace std;
const int MAXN = 5001;
vector< vector<int> > adj;
int cnt,low[MAXN],pre[MAXN],visit[MAXN],degree[MAXN];
void dfs(int u,int v)
{
visit[u]=1;
pre[u]=cnt++,low[u]=pre[u];
int i,len=adj[u].size();
for(i=0;i<len;i++){
if(adj[u][i]==v) continue;
if(!visit[adj[u][i]]) dfs(adj[u][i],u);
if(low[adj[u][i]]<low[u]) low[u]=low[adj[u][i]];
}
visit[u]=2;
}
int main(){
int i,j,u,v,n,m,len,ans;
scanf("%d %d",&n,&m);
adj.assign(n+1,vector<int>());
while(m--){
scanf("%d %d",&u,&v);
adj[u].push_back(v),adj[v].push_back(u);
}
memset(visit,0,sizeof(visit));
cnt=0,dfs(1,1);
memset(degree,0,sizeof(degree));
for(i=1;i<=n;i++){
len=adj[i].size();
for(j=0;j<len;j++)
if(low[i]!=low[adj[i][j]])
degree[low[i]]++;
}
for(ans=i=0;i<=n;i++)
if(degree[i]==1) ans++;
printf("%d\n",(ans+1)/2);
return 0;
}