Redundant Redundant Paths e-DCC path condensing point

Redundant Path Redundant Paths

Title transfer

sol:

If at least two non-overlapping paths exist between two points, indicating that they two components in communication with a double edge (cutting edge does not exist).

It can then be reduced to the point e-DCC, to give a tree.

For leaf nodes on the tree broad sense (of degree 1), still need at least one side connected to him.

So once even two leaf nodes can be greedy, the answer is obviously \ (cnt + 1 >> 1 \) .

#include<bits/stdc++.h>
#define IL inline
#define RG register
#define DB double
#define LL long long
using namespace std;

const int N=5005;
const int M=1e4+5;

int n,m,tot,cnt,leaf,Time,head[N],bel[N],vis[N],low[N],dfn[N],bri[M<<1];

struct edge{int x,y;}s[M];
struct EDGE{int next,to;}e[M<<1];

IL int gi() {
    RG int x=0,p=1; RG char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-') p=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') x=x*10+(ch^48),ch=getchar();
    return x*p;
}

IL void New() {
    tot=0;
    memset(&e,0,sizeof(e));
    memset(head,0,sizeof(head));
}

IL void make(int a,int b) {
    e[++tot]=(EDGE){head[a],b},head[a]=tot;
    e[++tot]=(EDGE){head[b],a},head[b]=tot;
}

void Tarjan(int x,int fx) {
    RG int i,y;
    dfn[x]=low[x]=++Time;
    for(i=head[x];i;i=e[i].next) {
        if(!dfn[y=e[i].to]) {
            Tarjan(y,x),low[x]=min(low[x],low[y]);
            if(low[y]>dfn[x]) bri[i]=bri[i^1]=1; 
        }
        else if(y!=fx) low[x]=min(low[x],dfn[y]);
    }   
}

void dfs(int x) {
    RG int i,y;
    bel[x]=cnt,vis[x]=1;
    for(i=head[x];i;i=e[i].next)
        if(!vis[y=e[i].to]&&!bri[i]) dfs(y);
}

void dfs2(int x,int fx) {
    RG int i,y,fl=0;
    for(i=head[x];i;i=e[i].next)
        if((y=e[i].to)!=fx) fl=1,dfs2(y,x);
    if(!fl) ++leaf;
}

int main()
{
    RG int i,b=0;
    n=gi(),m=gi(),tot=1;
    for(i=1;i<=m;++i)
        s[i].x=gi(),s[i].y=gi(),make(s[i].x,s[i].y);
    for(i=1,Tarjan(1,0);i<=n;++i)
        if(!vis[i]) ++cnt,dfs(i);
    if(cnt==1) return puts("0"),0;
    for(i=1,New();i<=m;++i)
        if(bel[s[i].x]!=bel[s[i].y]) make(bel[s[i].x],bel[s[i].y]);
    dfs2(1,0);
    for(i=head[1];i;i=e[i].next) ++b;
    if(b==1) ++leaf;
    printf("%d\n",leaf+1>>1);
    return 0;
}