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);
}

  

上一篇:css----transition的应用,产生动画效果。


下一篇:hdu 3352 求边双联通分量模板题(容器)