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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 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 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
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6 +---+---+ 4
/ 5
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7 +
Building new paths from 1 to 6 and from 4 to 7 satisfies the conditions. 1 2 3
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6 +---+---+ 4
/ 5 :
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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 |
题意：就是加入最少的边使之成双联通图
一个有桥的联通图要变成双联通图的化，先把双联通子图缩点处理，使之成为树（无环）
找出树的叶子节点（入度为1）加边的数为(leaf+1)/2;
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pair<int,int>pil;
const int maxn=5000+10;
const int maxm=10000+10;
struct node{
int to,next;
bool col;//为桥
}e[maxm];
int head[maxn],cnt;
int DFN[maxn],low[maxn];
int s[maxn],instack[maxn];
int idex,top,bridge;
int belong[maxn],in[maxn];
int n,m;
void init()
{
cnt=top=idex=bridge=0;
CLEAR(head,-1);
CLEAR(DFN,0);
CLEAR(low,0);
CLEAR(instack,0);
CLEAR(belong,0);
CLEAR(in,0);
}
void addedge(int u,int v)
{
e[cnt].to=v;e[cnt].next=head[u];
e[cnt].col=false;head[u]=cnt++;
}
void Tarjan(int u,int pre)
{
int v;
low[u]=DFN[u]=++idex;
s[top++]=u;
instack[u]=1;
for(int i=head[u];i!=-1;i=e[i].next)
{
v=e[i].to;
if(v==pre) continue;
if(!DFN[v])
{
Tarjan(v,u);
if(low[u]>low[v]) low[u]=low[v];
if(low[v]>DFN[u])//桥
{
bridge++;
e[i].col=true;
e[i^1].col=true;
}
}
else if(instack[v]&&low[u]>DFN[v])
low[u]=DFN[v];
}
if(low[u]==DFN[u])
{
cnt++;
do{
v=s[--top];
instack[v]=0;
belong[v]=cnt;
}while(v!=u);
}
}
void work()
{
REPF(i,1,n)
if(!DFN[i]) Tarjan(i,i);
for(int i=1;i<=n;i++)
{
for(int j=head[i];j!=-1;j=e[j].next)
if(e[j].col) in[belong[i]]++;
}
int ans=0;
REPF(i,1,cnt)
if(in[i]==1) ans++;
printf("%d\n",(ans+1)/2);
}
int main()
{
int u,v;
while(~scanf("%d%d",&n,&m))
{
init();
REPF(i,1,m)
{
scanf("%d%d",&u,&v);
addedge(u,v);
addedge(v,u);
}
work();
}
return 0;
}