POJ3352Road Construction

题目大意

给定一个无向连通图,问至少加几条边使得图成为边双联通图

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样例输入

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

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样例输出

Output for Sample Input 1
2

Output for Sample Input 2
0

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Solution

Tarjan将所有边双联通分量缩点以后,形成一棵树
答案就是(树上度数等于一的点的个数+1)>>1

#include
#include
#include
using namespace std;
#define maxn 1000010
inline int read(){
    int ret=0;
    char ch=getchar();
    while(ch<'0'||ch>'9') ch=getchar();
    while(ch>='0'&&ch<='9')ret=ret*10+ch-'0',ch=getchar();
    return ret;
}
struct Edge{
    int u,v,next;
}E[maxn];
int ecnt=0,head[maxn],dfn[maxn],low[maxn],deg[maxn],ord,vis[maxn];
void addedge(int u,int v){
    E[++ecnt].u=u;
    E[ecnt].v=v;
    E[ecnt].next=head[u];
    head[u]=ecnt;
}
void Tarjan(int x,int fx){
    dfn[x]=low[x]=++ord;
    vis[x]=1;
    for(int i=head[x];i;i=E[i].next){
        int v=E[i].v;
        if(v==fx) continue;
        if(dfn[v]==0) Tarjan(v,x),low[x]=min(low[x],low[v]);
        if(vis[v]==1) low[x]=min(dfn[v],low[x]);
    }
}
void init(){
    memset(E,0,sizeof(E));
    for(int i=0;i0;
    ecnt=0,ord=0;
}
int main(){
    //freopen("poj3352.in","r",stdin);
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        init();
        int ans=0;
        for(int i=1;i<=m;++i){
            int aa=read(),bb=read();
            addedge(aa,bb),addedge(bb,aa);
        }
        for(int i=1;i<=n;++i){
            if(dfn[i]==0) Tarjan(i,0);
        }
        for(int i=1;i<=ecnt;++i){
            if(low[E[i].u]!=low[E[i].v])++deg[low[E[i].u]];
        }
        for(int i=1;i<=n;++i) if(deg[i]==1) ++ans;
        printf("%d\n",(ans+1)>>1);
    }
    return 0;
}