ACM学习-割点和桥

// ACM学习-割点和桥.cpp : 定义控制台应用程序的入口点。

//

#include "stdafx.h"

#include<iostream>

using namespace std;

const int v = 13;

int edge[v][v] = {
{0,1,1,0,0,1,0,0,0,0,0,1,0},
{1,0,1,1,0,0,0,1,0,0,0,0,1},
{1,1,0,0,0,0,0,0,0,0,0,0,0},
{0,1,0,0,1,0,0,0,0,0,0,0,0},
{0,0,0,1,0,0,0,0,0,0,0,0,0},
{1,0,0,0,0,0,0,0,0,0,0,0,0},
{0,1,0,0,0,0,0,1,1,0,1,0,0},
{0,1,0,0,0,0,1,0,0,0,1,0,0},
{0,0,0,0,0,0,1,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,1,1},
{0,0,0,0,0,0,1,1,0,0,0,0,0},
{1,0,0,0,0,0,0,0,0,1,0,0,1},
{0,1,0,0,0,0,0,0,0,1,0,1,0}

};

bool bridge[v][v], cut[v];

int low[v] = {0}, dfn[v] = { 0 }, vis[v] = { 0 };

void cut_bridge(int cur,int father,int dep,int n){
vis[cur] = 1; dfn[cur] = low[cur] = dep;
int children = 0;
for (int i = 0; i < n; i++){
if (edge[cur][i]){
if (i != father && 1 == vis[i]){
//看看周围有没有更快到达祖先的点，有的话减少low[cur]                           
if (dfn[i] < low[cur])
low[cur] = dfn[i];
}
if (0 == vis[i]){//没有访问过，叶子节点肯定不访问该段代码块
cut_bridge(i,cur,dep+1,n);
children++;
//儿子更快到达祖先，改善一下爸爸的low[i];
if (low[i] < low[cur]) low[cur] = low[i];
//如果跟节点有两个孩子，则它是关键节点
//如果子节点，没有周围的节点改善它到祖先的深度，即只有一个父节点
if ((father==-1&&children>1||(father!=-1&&low[i]>=dfn[cur]))){
cut[cur] = true;
cout << "关键节点：" << cur << endl;
}
if (low[i] > dfn[cur])
{ 
bridge[cur][i] = true;
cout << "桥是：" << cur << endl;
}
}
}
}
vis[cur] = 2;

}

int _tmain(int argc, _TCHAR* argv[])

{

cut_bridge(0,-1,1,13);
return 0;

}