poj3177 无向连通图加多少条边变成边双连通图

Redundant Paths

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15752		Accepted: 6609

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
注意重边
不能通过low值判断两个点在一个连通分量内，
因为这样的数据 1-6都在一个联通块内，但6的low值和其他不同。
6 7
1 2
2 3
3 4
4 6
6 2
4 5
5 1
但是貌似 没有影响？。因为虽然6的low值有变化，但是他不可能是叶节点，所以用low值水过去也是可以的吧


就是缩点后找叶子结点的数目+1/2就是答案

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=5050;
bool mp[N][N];
int head[N],dfn[N],low[N],cont[N];
int tot,cnt;
struct node{
   int next,to;
}e[N<<1];
void add(int u,int v){
   e[tot].to=v;e[tot].next=head[u];head[u]=tot++;
   e[tot].to=u;e[tot].next=head[v];head[v]=tot++;
}
void Tajan(int u,int fa){
   dfn[u]=low[u]=++cnt;
   for(int i=head[u];i+1;i=e[i].next){
    int v=e[i].to;
    if(v==fa) continue;
    if(!dfn[v]) {
        Tajan(v,u);
        low[u]=min(low[v],low[u]);
    }
    else low[u]=min(low[u],dfn[v]);
   }
}
int main(){
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        int u,v;
        memset(head,-1,sizeof(head));
        memset(mp,0,sizeof(mp));
        memset(cont,0,sizeof(cont));
        tot=cnt=0;
        while(m--){
            scanf("%d%d",&u,&v);
            if(mp[u][v]) continue;
            add(u,v);
            mp[u][v]=mp[v][u]=1;
        }
        Tajan(1,0);
        for(int i=1;i<=n;++i)
        for(int j=head[i];j+1;j=e[j].next){
            int v=e[j].to;
            if(low[v]!=low[i])
                ++cont[low[i]];
        }
        int ans=0;
        for(int i=1;i<=n;++i)
            if(cont[i]==1) ++ans;
            printf("%d
",(ans+1)/2);
    }
}

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=5050;
bool mp[N][N];
int head[N],dfn[N],low[N],cont[N],bl[N],q[N];
int tot,cnt,scnt,l;
struct node{
   int next,to;
}e[N<<1];
void add(int u,int v){
   e[tot].to=v;e[tot].next=head[u];head[u]=tot++;
   e[tot].to=u;e[tot].next=head[v];head[v]=tot++;
}
void Tajan(int u,int fa){
    q[l++]=u;
   dfn[u]=low[u]=++cnt;
   for(int i=head[u];i+1;i=e[i].next){
    int v=e[i].to;
    if(v==fa) continue;
    if(!dfn[v]) {
        Tajan(v,u);
        low[u]=min(low[v],low[u]);
    }
    else low[u]=min(low[u],dfn[v]);
   }
   if(dfn[u]==low[u]){
    int t;
    ++scnt;
    do{
        t=q[--l];
        bl[t]=scnt;
    }while(t!=u);
   }
}
int main(){
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        int u,v;
        memset(head,-1,sizeof(head));
        memset(mp,0,sizeof(mp));
        memset(cont,0,sizeof(cont));
        l=scnt=tot=cnt=0;
        while(m--){
            scanf("%d%d",&u,&v);
            if(mp[u][v]) continue;
            add(u,v);
            mp[u][v]=mp[v][u]=1;
        }
        Tajan(1,0);
        for(int i=1;i<=n;++i)
        for(int j=head[i];j+1;j=e[j].next){
            int v=e[j].to;
            if(bl[v]!=bl[i])
                ++cont[bl[i]];
        }
        int ans=0;
        for(int i=1;i<=n;++i)
            if(cont[i]==1) ++ans;
            printf("%d
",(ans+1)/2);
    }
}