边双连通分量
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
const int N = 5e4+10, M = 3e5+10;
int h[N],e[M*2],ne[M*2],idx;
int dfn[N],low[N],c[N];
bool bridge[M*2];
int n,m,num,dcc;
void add(int a,int b){
e[++idx] = b,ne[idx] = h[a], h[a] = idx;
}
void tarjan(int x,int in_edge){
dfn[x] = low[x] = ++num;
for(int i=h[x];i;i=ne[i]){
int y = e[i];
if(!dfn[y]){
tarjan(y,i);
low[x] = min(low[x],low[y]);
if(low[y] > dfn[x])
bridge[i] = bridge[i ^ 1] = true;
}
else if(i != (in_edge ^ 1)){
low[x] = min(low[x],dfn[y]);
}
}
}
void dfs(int x){
c[x] = dcc;
for(int i = h[x];i;i = ne[i]){
int y = e[i];
if(c[y] || bridge[i]) continue;
dfs(y);
}
}
int main(){
cin >> n >> m;
idx = 1;
for(int i=1;i<=m;i++){
int x,y;
scanf("%d%d",&x,&y);
add(x,y), add(y,x);
}
for(int i=1;i<=n;i++)
if(!dfn[i]) tarjan(i,0);
for(int i=1;i<=n;i++)
if(!c[i]) ++dcc, dfs(i);
printf("There are %d e-DCCs.\n",dcc);
for(int i=1;i<=n;i++)
printf("%dbelongs to DCC %d.\n",i,c[i]);
return 0;
}
e-DCC的缩点:
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
const int N = 5e4+10, M = 3e5+10;
int h[N],e[M*2],ne[M*2],idx;
int hc[N],ec[M*2],nec[M*2],idxc;
int dfn[N],low[N],c[N];
bool bridge[M*2];
int n,m,num,dcc;
void add(int a,int b){
e[++idx] = b,ne[idx] = h[a], h[a] = idx;
}
void add_c(int a,int b){
ec[++idxc] = b,nec[idxc] = hc[a] , hc[a] = idxc;
}
void tarjan(int x,int in_edge){
dfn[x] = low[x] = ++num;
for(int i=h[x];i;i=ne[i]){
int y = e[i];
if(!dfn[y]){
tarjan(y,i);
low[x] = min(low[x],low[y]);
if(low[y] > dfn[x])
bridge[i] = bridge[i ^ 1] = true;
}
else if(i != (in_edge ^ 1)){
low[x] = min(low[x],dfn[y]);
}
}
}
void dfs(int x){
c[x] = dcc;
for(int i = h[x];i;i = ne[i]){
int y = e[i];
if(c[y] || bridge[i]) continue;
dfs(y);
}
}
int main(){
cin >> n >> m;
idx = 1;
idxc = 1;
for(int i=1;i<=m;i++){
int x,y;
scanf("%d%d",&x,&y);
add(x,y), add(y,x);
}
for(int i=1;i<=n;i++)
if(!dfn[i]) tarjan(i,0);
for(int i=1;i<=n;i++)
if(!c[i]) ++dcc, dfs(i);
for(int i=2;i<=idx;i++){
int x = e[i^1], y = e[i];
if(c[x] == c[y]) continue;
add_c(c[x],c[y]);
}
printf("缩点之后的森林,点数%d(可能有重边)\n",dcc,idxc / 2);
for(int i=2;i<idxc;i+=2)
printf("%d %d\n",ec[i^1],ec[i]);
return 0;
}
点双连通分量:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
const int N = 5e4+10, M = 3e5+10;
int h[N],e[M*2],ne[M*2],idx;
int hc[N],ec[M*2],nec[M*2],idxc;
int dfn[N],low[N],c[N];
vector<int> dcc[N];
bool cut[N];
int n,m,num,cnt,root;
int stack[N],top;
int new_id[N];
void add(int a,int b){
e[++idx] = b,ne[idx] = h[a],h[a] = idx;
}
void add_c(int a,int b){
ec[++idxc] = b,nec[idxc] = hc[a], hc[a] = idxc;
}
void tarjan(int x){
dfn[x] = low[x] = ++num;
stack[++top] = x;
if(x == root && h[x] == 0){
dcc[++cnt].push_back(x);
return;
}
int flag = 0;
for(int i=h[x];i;i=ne[i]){
int y = e[i];
if(!dfn[y]){
tarjan(y);
low[x] = min(low[x],low[y]);
if(low[y] >= dfn[x]){
flag++;
if(x != root || flag > 1) cut[x] = true;
cnt++;
int z;
do{
z = stack[top--];
dcc[cnt].push_back(z);
}while(z != y);
dcc[cnt].push_back(x);
}
}
else low[x] = min(low[x],dfn[y]);
}
for(int i=1;i<=cnt;i++){
}
}
int main(){
cin >> n >> m;
idx = 1;
for(int i=1;i<=m;i++){
int x,y;
scanf("%d%d",&x,&y);
add(x,y),add(y,x);
}
for(int i=1;i<=n;i++)
if(!dfn[i]) root = i,tarjan(i);
for(int i=1;i<=cnt;i++){
printf("v-DCC #%d:",i);
for(int j = 0;j<dcc[i].size();j++)
printf(" %d ",dcc[i][j]);
puts(" ");
}
return 0;
}
点双连通分量缩点:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
const int N = 5e4+10, M = 3e5+10;
int h[N],e[M*2],ne[M*2],idx;
int hc[N],ec[M*2],nec[M*2],idxc;
int dfn[N],low[N],c[N];
vector<int> dcc[N];
bool cut[N];
int n,m,num,cnt,root;
int stack[N],top;
int new_id[N];
void add(int a,int b){
e[++idx] = b,ne[idx] = h[a],h[a] = idx;
}
void add_c(int a,int b){
ec[++idxc] = b,nec[idxc] = hc[a], hc[a] = idxc;
}
void tarjan(int x){
dfn[x] = low[x] = ++num;
stack[++top] = x;
if(x == root && h[x] == 0){
dcc[++cnt].push_back(x);
return;
}
int flag = 0;
for(int i=h[x];i;i=ne[i]){
int y = e[i];
if(!dfn[y]){
tarjan(y);
low[x] = min(low[x],low[y]);
if(low[y] >= dfn[x]){
flag++;
if(x != root || flag > 1) cut[x] = true;
cnt++;
int z;
do{
z = stack[top--];
dcc[cnt].push_back(z);
}while(z != y);
dcc[cnt].push_back(x);
}
}
else low[x] = min(low[x],dfn[y]);
}
for(int i=1;i<=cnt;i++){
}
}
int main(){
cin >> n >> m;
idx = 1;
for(int i=1;i<=m;i++){
int x,y;
scanf("%d%d",&x,&y);
add(x,y),add(y,x);
}
for(int i=1;i<=n;i++)
if(!dfn[i]) root = i,tarjan(i);
num = cnt;
for(int i=1;i<=n;i++)
if(cut[i]) new_id[i] = ++num;
idxc = 1;
for(int i=1;i<=cnt;i++){
for(int j=0;j<dcc[i].size();j++){
int x = dcc[i][j];
if(cut[x]){
add_c(i,new_id[x]);
add_c(new_id[x],i);
}
else c[x] = i;
}
}
printf("缩点之后的森林,点数 %d, 边数 %d\n",num,idxc / 2);
printf("编号1~%d 的为原图的V-DCC,编号 >%d 的为原图割点\n",cnt,cnt);
for(int i=2;i<idxc;i+=2)
printf("%d %d\n",ec[i^1],ec[i]);
return 0;
}