Description
In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of often being forced to take a particular path and want to build some new paths so that they will always have a choice of at least two separate routes between any pair of fields. They currently have at least one route between each pair of fields and want to have at least two. Of course, they can only travel on Official Paths when they move from one field to another.
Given a description of the current set of R (F-1 <= R <= 10,000) paths that each connect exactly two different fields, determine the minimum number of new paths (each of which connects exactly two fields) that must be built so that there are at least two separate routes between any pair of fields. Routes are considered separate if they use none of the same paths, even if they visit the same intermediate field along the way.
There might already be more than one paths between the same pair of fields, and you may also build a new path that connects the same fields as some other path.
Input
Line 1: Two space-separated integers: F and R
Lines 2..R+1: Each line contains two space-separated integers which are the fields at the endpoints of some path.
Output
Line 1: A single integer that is the number of new paths that must be built.
Sample Input
7 7
1 2
2 3
3 4
2 5
4 5
5 6
5 7
Sample Output
2
Hint
Explanation of the sample:
One visualization of the paths is:
1 2 3
+---+---+
| |
| |
6 +---+---+ 4
/ 5
/
/
7 +
Building new paths from 1 to 6 and from 4 to 7 satisfies the conditions.
1 2 3
+---+---+
: | |
: | |
6 +---+---+ 4
/ 5 :
/ :
/ :
7 + - - - -
Check some of the routes:
1 – 2: 1 –> 2 and 1 –> 6 –> 5 –> 2
1 – 4: 1 –> 2 –> 3 –> 4 and 1 –> 6 –> 5 –> 4
3 – 7: 3 –> 4 –> 7 and 3 –> 2 –> 5 –> 7 Every pair of fields is, in fact, connected by two routes.
It's possible that adding some other path will also solve the problem (like one from 6 to 7). Adding two paths, however, is the minimum.
Source
題意大意：一群牛將被在一個(gè)特定路徑構(gòu)成的農(nóng)場(chǎng)上遷移，每?jī)蓧K農(nóng)場(chǎng)之間都至少有一條通道，這些牛要求每?jī)蓧K路徑至少要有兩條通道，求最少需要修建多少條路才能滿足要求。
這題的解法與http://m.shnenglu.com/mythit/archive/2009/05/29/86082.html完全一樣，只是題目中說了圖中有可能存在平行邊，這里必須判斷一下。我還是很偷懶的用了STL里的vector模擬鄰接矩陣，并且開了個(gè)5001*5001的bool數(shù)組判斷平行邊。結(jié)果導(dǎo)致代碼的效率和空間消耗都很大，110MS和將近24M的內(nèi)存空間。如果自己建圖的話，效率能提高很多。
#include <iostream>
#include <vector>
using namespace std;
const int MAXN = 5001;
vector< vector<int> > adj;
bool hash[MAXN][MAXN];
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;
while(scanf("%d %d",&n,&m)!=EOF){
adj.assign(n+1,vector<int>());
memset(hash,false,sizeof(hash));
while(m--){
scanf("%d %d",&u,&v);
if(!hash[u][v]){
hash[u][v]=true;
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;
}