边双连通模版
只有一个边双连通：{1,2,3,4,5}
去掉桥，每块缩为一个点
调用方法：
init();
solve(int l, int r){}; [l, r] 是点标
suodian();
双连通 缩点 求桥模版：
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <queue>
#include <vector>
using namespace std;
#define N 100005
#define M 200300
#define inf 10000000
struct Edge{
int from,to,next;
bool cut;
}edge[2*M];
int head[N],edgenum;
int Low[N],DFN[N],Stack[N];//Belong数组的值是1~block
int Index,top;
int Belong[N],block;//新图的连通块标号（1~block）
bool Instack[N];
int bridge; //割桥数量
void addedge(int u,int v){
Edge E={u,v,head[u],0}; edge[edgenum]=E; head[u] = edgenum++;
Edge E2={v,u,head[v],0};edge[edgenum]=E2;head[v] = edgenum++;
}
void Tarjan(int u,int pre){
int v;
Low[u] = DFN[u] = ++Index;
Stack[top++] = u;
Instack[u] = true;
for(int i = head[u]; ~i ;i = edge[i].next){
v = edge[i].to;
// 如果重边有效的话下面这句改成: if(v == pre && pre_num == 0){pre_num++;continue;} pre_num在for上面定义 int pre_num=0;
if( v == pre )continue;
if( !DFN[v] ){
Tarjan(v,u);
Low[u] = min(Low[u], Low[v]);
if(Low[v] > DFN[u]){
bridge++;
edge[i].cut = true;
edge[i^1].cut = true;
}
}
else if(Instack[v])Low[u] = min(Low[u], DFN[v]);
}
if(Low[u] == DFN[u]){
block++;
do{
v = Stack[--top];
Instack[v] = false;
Belong[v] = block;
}while( v != u );
}
}
void work(int l, int r){
memset(DFN,0,sizeof(DFN));
memset(Instack,false,sizeof(Instack));
Index = top = block = bridge = 0;
for(int i = l; i <= r; i++)if(!DFN[i])Tarjan(i,i);
}
vector<int>G[N];//点标从1-block
void suodian(){
for(int i = 1; i <= block; i++)G[i].clear();
for(int i = 0; i < edgenum; i+=2){
int u = Belong[edge[i].from], v = Belong[edge[i].to];
if(u==v)continue;
G[u].push_back(v), G[v].push_back(u);
}
}
void init(){edgenum = 0; memset(head,-1,sizeof(head));}
求割点和桥 (binshen模版）
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int INF = 0x3f3f3f3f;
/*
* 求 无向图的割点和桥
* 可以找出割点和桥，求删掉每个点后增加的连通块。
* 需要注意重边的处理，可以先用矩阵存，再转邻接表，或者进行判重
*/
#define N 10005
#define M 200010
struct Edge{
int to,next,w;
bool cut;//是否为桥的标记
}edge[M*2];
int head[N],edgenum;
int Low[N],DFN[N],Stack[N],Index,top;
bool Instack[N];
bool cut[N]; //是否为割点
int add_block[N];//删除i点后增加的连通块数量为add_block[i]
int bridge; //桥数量
void addedge(int u,int v,int w)
{
edge[edgenum].to = v;edge[edgenum].next = head[u];edge[edgenum].cut = false;
edge[edgenum].w = w;
head[u] = edgenum++;
edge[edgenum].to = u;edge[edgenum].next = head[v];edge[edgenum].cut = false;
edge[edgenum].w = w;
head[v] = edgenum++;
}
void Tarjan(int u,int pre){
int v;
Low[u] = DFN[u] = ++Index;
Stack[top++] = u;
Instack[u] = true;
int son = 0;
int pre_num = 0;
for(int i = head[u];i != -1;i = edge[i].next)
{
v = edge[i].to;
//若重边无效则把下面这句换成 if(v == pre)continue;
if(v == pre && pre_num == 0){pre_num++;continue;}
if( !DFN[v] )
{
son++;
Tarjan(v,u);
if(Low[u] > Low[v])Low[u] = Low[v];
//桥
//一条无向边(u,v)是桥，当且仅当(u,v)为树枝边，且满足DFS(u)<Low(v)。
if(Low[v] > DFN[u])
{
bridge++;
edge[i].cut = true;
edge[i^1].cut = true;
}
//割点
//一个顶点u是割点，当且仅当满足(1)或(2) (1) u为树根，且u有多于一个子树。
//(2) u不为树根，且满足存在(u,v)为树枝边(或称父子边，
//即u为v在搜索树中的父亲)，使得DFS(u)<=Low(v)
if(u != pre && Low[v] >= DFN[u])//不是树根
{
cut[u] = true;
add_block[u]++;
}
}
else if( Low[u] > DFN[v])
Low[u] = DFN[v];
}
//树根，分支数大于1
if(u == pre && son > 1)cut[u] = true;
if(u == pre)add_block[u] = son - 1;
Instack[u] = false;
top--;
}
void solve(int l, int r){ //点标区间为[l,r]
memset(DFN,0,sizeof(DFN));
memset(Instack,false,sizeof(Instack));
memset(add_block,0,sizeof(add_block));
memset(cut,false,sizeof(cut));
Index = top = bridge = 0;
for(int i = l;i <= r;i++)if(!DFN[i])Tarjan(i,i);
}
void init(){ edgenum = 0; memset(head,-1,sizeof(head));}
模版1：重边有效 白书模版：
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<vector>
#include<math.h>
#include<queue>
#include<set>
using namespace std;
#define N 10050
#define M 200005
int n,m;//n个点 m条边 点标从1开始
struct Edge{
int from,to,val,nex;
bool cut;//记录这条边是否为割边
}edge[2*M];//双向边则 edge[i]与edge[i^1]是2条反向边
int head[N],edgenum;//在一开始就要 memset(head,-1,sizeof(head)); edgenum=0;
int low[N],dfn[N],tarjin_time;
void add(int u,int v,int w){
Edge E={u,v,w,head[u],0};
edge[edgenum]=E;
head[u]=edgenum++;
Edge E2={v,u,w,head[v],0};
edge[edgenum]=E2;
head[v]=edgenum++;
}
void tarjin(int u,int pre)
{
low[u]=dfn[u]= ++tarjin_time;
int flag=1;//flag是阻止双向边的反向边 i和i^1
for(int i=head[u];i!=-1;i=edge[i].nex)
{
int v=edge[i].to;
if(flag&&v==pre)
{
flag=0;
continue;
}
if(!dfn[v])
{
tarjin(v,u);
if(low[u]>low[v])low[u]=low[v];
if(low[v]>dfn[u])//是桥low[v]表示v能走到的最早祖先 有重边且u是v的最早祖先 则low[v]==dfn[u],所以不会当作桥
edge[i].cut=edge[i^1].cut=true;
}
else if(low[u]>dfn[v])low[u]=dfn[v];
}
}
void find_edgecut()
{
memset(dfn,0,sizeof(dfn));
tarjin_time=0;
for(int i=1;i<=n;i++)if(!dfn[i])tarjin(i,i);
}
void init(){memset(head, -1, sizeof head); edgenum = 0;}
重边算1条：
#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <vector>
using namespace std;
/*
* 求 无向图的割点和桥
* 可以找出割点和桥，求删掉每个点后增加的连通块。
* 这个模版是重边算1条
*/
const int MAXN = 10010;
const int MAXM = 100010;
struct Edge{
int to,next;
bool cut;//是否为桥的标记
}edge[MAXM*2];
int head[MAXN],edgenum;
int Low[MAXN],DFN[MAXN],Stack[MAXN];
int Index,top;
bool Instack[MAXN];
bool cut[MAXN]; //点是否为割点
int add_block[MAXN];//删除一个点后增加的连通块
int bridge;
void add(int u,int v){
edge[edgenum].to = v;edge[edgenum].next = head[u];edge[edgenum].cut = false;
head[u] = edgenum++;
edge[edgenum].to = u;edge[edgenum].next = head[v];edge[edgenum].cut = false;
head[v] = edgenum++;
}
void Tarjan(int u,int pre){
int v;
Low[u] = DFN[u] = ++Index;
Stack[top++] = u;
Instack[u] = true;
int son = 0;
for(int i = head[u];i != -1;i = edge[i].next)
{
v = edge[i].to;
if(v == pre)continue;
if( !DFN[v] )
{
son++;
Tarjan(v,u);
if(Low[u] > Low[v])Low[u] = Low[v];
//桥
//一条无向边(u,v)是桥，当且仅当(u,v)为树枝边，且满足DFS(u)<Low(v)。
if(Low[v] > DFN[u])
{
bridge++;
edge[i].cut = true;
edge[i^1].cut = true;
}
//割点
//一个顶点u是割点，当且仅当满足(1)或(2) (1) u为树根，且u有多于一个子树。
//(2) u不为树根，且满足存在(u,v)为树枝边(或称父子边，
//即u为v在搜索树中的父亲)，使得DFS(u)<=Low(v)
if(u != pre && Low[v] >= DFN[u])//不是树根
{
cut[u] = true;
add_block[u]++;
}
}
else if( Low[u] > DFN[v])
Low[u] = DFN[v];
}
//树根，分支数大于1
if(u == pre && son > 1)cut[u] = true;
if(u == pre)add_block[u] = son - 1;
Instack[u] = false;
top--;
}
void solve(int n){
memset(DFN,0,sizeof(DFN));
memset(Instack,false,sizeof(Instack));
memset(add_block,0,sizeof(add_block));
memset(cut,false,sizeof(cut));
Index = top = 0;
bridge = 0;
for(int i = 1;i <= n;i++)if(!DFN[i]) Tarjan(i,i);
}
void init(){edgenum = 0; memset(head,-1,sizeof(head));}