POJ3177 & 求边双联通分量

题意:

  给一张无向图,求加多少边使原图任意两点边双联通.

SOL:

  一个不会写边双点双强联通的傻逼.

  一个结论:把一棵树变成满足条件的图需要加的边使入度为1的点数+1除以2.------>就是树的叶子两两连上.

  然后就是缩点然后统计就好了...

  然后学会了怎么搞边双...看了几天白书还是非常显然的感觉...

Code:

  

/*==========================================================================
# Last modified: 2016-03-13 17:55
# Filename: poj3177.cpp
# Description: 
==========================================================================*/
#define me AcrossTheSky 
#include <cstdio> 
#include <cmath> 
#include <ctime> 
#include <string> 
#include <cstring> 
#include <cstdlib> 
#include <iostream> 
#include <algorithm> 
  
#include <set> 
#include <map> 
#include <stack> 
#include <queue> 
#include <vector> 
 
#define lowbit(x) (x)&(-x) 
#define FOR(i,a,b) for((i)=(a);(i)<=(b);(i)++) 
#define FORP(i,a,b) for(int i=(a);i<=(b);i++) 
#define FORM(i,a,b) for(int i=(a);i>=(b);i--) 
#define ls(a,b) (((a)+(b)) << 1) 
#define rs(a,b) (((a)+(b)) >> 1) 
#define getlc(a) ch[(a)][0] 
#define getrc(a) ch[(a)][1] 
 
#define maxn 100000 
#define maxm 100000 
#define pi 3.1415926535898 
#define _e 2.718281828459 
#define INF 1070000000 
using namespace std; 
typedef long long ll; 
typedef unsigned long long ull; 
 
template<class T> inline 
void read(T& num) { 
    bool start=false,neg=false; 
    char c; 
    num=0; 
    while((c=getchar())!=EOF) { 
        if(c=='-') start=neg=true; 
        else if(c>='0' && c<='9') { 
            start=true; 
            num=num*10+c-'0'; 
        } else if(start) break; 
    } 
    if(neg) num=-num; 
} 
/*==================split line==================*/
int to[maxm],next[maxm],first[maxn];
int low[maxn],dfn[maxn],s[maxn],belong[maxn],in[maxn];
bool instack[maxn];
int sume=1,scc,clo=0,top,sum=0,n,m;
void addedge(int x,int y){
	sume++; to[sume]=y; next[sume]=first[x]; first[x]=sume;
}
void tarjan(int u,int fa){
	low[u]=dfn[u]=(++clo);
	instack[u]=1;
	s[++top]=u;
	for (int i=first[u];i;i=next[i]){
		if (i==(fa^1)) continue;
		int v=to[i];
		if (!dfn[v]){
			tarjan(v,i);
			low[u]=min(low[u],low[v]);
		}
		else if (instack[v]) low[u]=min(low[u],dfn[v]);
	}
	if (dfn[u]==low[u]){
		scc++;
		while (true){
			int v=s[top--];
			instack[v]=false;
			belong[v]=scc;
			if (v==u) break;
		}
	}
}
int main(){
	read(n); read(m);
	FORP(i,1,m){
		int x,y; 
		read(x); read(y);
		addedge(x,y); addedge(y,x);
	}
	FORP(i,1,n) 
		if (!dfn[i]) tarjan(1,0);
	FORP(i,1,n)
		for (int j=first[i];j;j=next[j])
			if (belong[i]!=belong[to[j]]) in[belong[i]]++;
	int sum=0;
	FORP(i,1,n) if (in[i]==1) sum++;
	printf("%d\n",(sum+1)/2);
}

 

  

posted @ 2016-03-13 19:12  YCuangWhen  阅读(201)  评论(0编辑  收藏  举报