【专题】图的连通性问题
有向图的强连通分量
POJ 1236 - Network of Schools(基础)
http://acm.pku.edu.cn/JudgeOnline/problem?id=1236
题意：问添加多少边可成为完全连通图
解法：缩点，看度数
/** head-file **/
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <algorithm>
/** define-for **/
#define REP(i, n) for (int i=0;i<int(n);++i)
#define FOR(i, a, b) for (int i=int(a);i<int(b);++i)
#define DWN(i, b, a) for (int i=int(b-1);i>=int(a);--i)
#define REP_1(i, n) for (int i=1;i<=int(n);++i)
#define FOR_1(i, a, b) for (int i=int(a);i<=int(b);++i)
#define DWN_1(i, b, a) for (int i=int(b);i>=int(a);--i)
#define REP_N(i, n) for (i=0;i<int(n);++i)
#define FOR_N(i, a, b) for (i=int(a);i<int(b);++i)
#define DWN_N(i, b, a) for (i=int(b-1);i>=int(a);--i)
#define REP_1_N(i, n) for (i=1;i<=int(n);++i)
#define FOR_1_N(i, a, b) for (i=int(a);i<=int(b);++i)
#define DWN_1_N(i, b, a) for (i=int(b);i>=int(a);--i)
/** define-useful **/
#define clr(x,a) memset(x,a,sizeof(x))
#define sz(x) int(x.size())
#define see(x) cerr<<#x<<" "<<x<<endl
#define se(x) cerr<<" "<<x
/** test **/
#define Display(A, n, m) { \
REP(i, n){ \
REP(j, m) cout << A[i][j] << " "; \
cout << endl; \
} \
}
#define Display_1(A, n, m) { \
REP_1(i, n){ \
REP_1(j, m) cout << A[i][j] << " "; \
cout << endl; \
} \
}
using namespace std;
/** typedef **/
typedef long long LL;
/** Add - On **/
const int direct4[4][2]={ {0,1},{1,0},{0,-1},{-1,0} };
const int direct8[8][2]={ {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int direct3[6][3]={ {1,0,0},{0,1,0},{0,0,1},{-1,0,0},{0,-1,0},{0,0,-1} };
const int MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long INFF = 1LL << 60;
const double EPS = 1e-9;
const double OO = 1e15;
const double PI = acos(-1.0); //M_PI;
const int maxn=111111;
const int maxm=511111;
int n,m;
struct EDGENODE{
int to;
int w;
int next;
};
struct SGRAPH{
int head[maxn];
EDGENODE edges[maxm];
int edge;
void init()
{
clr(head,-1);
edge=0;
}
void addedge(int u,int v,int c=1)
{
edges[edge].w=c,edges[edge].to=v,edges[edge].next=head[u],head[u]=edge++;
}
int pre[maxn],lowlink[maxn],sccno[maxn],scc_cnt,dfs_clock;
stack<int>stk;
void dfs(int u)
{
pre[u]=lowlink[u]=++dfs_clock;
stk.push(u);
for (int i=head[u];i!=-1;i=edges[i].next){
int v=edges[i].to;
if (!pre[v]){
dfs(v);
lowlink[u]=min(lowlink[u],lowlink[v]);
} else if (!sccno[v]){
lowlink[u]=min(lowlink[u],pre[v]);
}
}
if (lowlink[u]==pre[u]){
scc_cnt++;
int x;
do{
x=stk.top();
stk.pop();
sccno[x]=scc_cnt;
}while (x!=u);
}
}
void find_scc(int n)
{
dfs_clock=scc_cnt=0;
clr(sccno,0);
clr(pre,0);
while (!stk.empty()) stk.pop();
REP_1(i,n) if (!pre[i]) dfs(i);
}
};
SGRAPH solver;
bool mp[111][111];
int main()
{
int ans1,ans2;
int idd,odd;
while (~scanf("%d",&n))
{
clr(mp,0);
solver.init();
REP_1(i,n)
{
int xt;
while (~scanf("%d",&xt))
{
if (xt==0) break;
solver.addedge(i,xt);
}
}
solver.find_scc(n);
REP(u,n)
{
for (int i=solver.head[u];i!=-1;i=solver.edges[i].next)
{
int v=solver.edges[i].to;
if (solver.sccno[u]!=solver.sccno[v])
{
mp[solver.sccno[u]][solver.sccno[v]]=true;
}
}
}
//Display_1(mp,solver.scc_cnt,solver.scc_cnt);
ans1=ans2=0;
m=solver.scc_cnt;
REP_1(i,m)
{
idd=0;
odd=0;
REP_1(j,m)
{
if (mp[j][i]) idd++;
if (mp[i][j]) odd++;
}
if (!idd) ans1++;
if (!odd) ans2++;
}
ans2=max(ans1,ans2);
if (m==1) ans2=0;
printf("%d\n%d\n",ans1,ans2);
}
return 0;
}
POJ 2553 - The Bottom of a Graph(基础)
找出度为零的强连通分量，把符合条件的点都输出出来。
/** head-file **/
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <algorithm>
/** define-for **/
#define REP(i, n) for (int i=0;i<int(n);++i)
#define FOR(i, a, b) for (int i=int(a);i<int(b);++i)
#define DWN(i, b, a) for (int i=int(b-1);i>=int(a);--i)
#define REP_1(i, n) for (int i=1;i<=int(n);++i)
#define FOR_1(i, a, b) for (int i=int(a);i<=int(b);++i)
#define DWN_1(i, b, a) for (int i=int(b);i>=int(a);--i)
#define REP_N(i, n) for (i=0;i<int(n);++i)
#define FOR_N(i, a, b) for (i=int(a);i<int(b);++i)
#define DWN_N(i, b, a) for (i=int(b-1);i>=int(a);--i)
#define REP_1_N(i, n) for (i=1;i<=int(n);++i)
#define FOR_1_N(i, a, b) for (i=int(a);i<=int(b);++i)
#define DWN_1_N(i, b, a) for (i=int(b);i>=int(a);--i)
/** define-useful **/
#define clr(x,a) memset(x,a,sizeof(x))
#define sz(x) int(x.size())
#define see(x) cerr<<#x<<" "<<x<<endl
#define se(x) cerr<<" "<<x
#define pb push_back
#define mp make_pair
/** test **/
#define Display(A, n, m) { \
REP(i, n){ \
REP(j, m) cout << A[i][j] << " "; \
cout << endl; \
} \
}
#define Display_1(A, n, m) { \
REP_1(i, n){ \
REP_1(j, m) cout << A[i][j] << " "; \
cout << endl; \
} \
}
using namespace std;
/** typedef **/
typedef long long LL;
/** Add - On **/
const int direct4[4][2]={ {0,1},{1,0},{0,-1},{-1,0} };
const int direct8[8][2]={ {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int direct3[6][3]={ {1,0,0},{0,1,0},{0,0,1},{-1,0,0},{0,-1,0},{0,0,-1} };
const int MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long INFF = 1LL << 60;
const double EPS = 1e-9;
const double OO = 1e15;
const double PI = acos(-1.0); //M_PI;
const int maxn=111111;
const int maxm=511111;
int n,m;
struct EDGENODE{
int to;
int w;
int next;
};
struct SGRAPH{
int head[maxn];
EDGENODE edges[maxm];
int edge;
void init()
{
clr(head,-1);
edge=0;
}
void addedge(int u,int v,int c=0)
{
edges[edge].w=c,edges[edge].to=v,edges[edge].next=head[u],head[u]=edge++;
}
int pre[maxn],lowlink[maxn],sccno[maxn],scc_cnt,dfs_clock;
stack<int>stk;
void dfs(int u)
{
pre[u]=lowlink[u]=++dfs_clock;
stk.push(u);
for (int i=head[u];i!=-1;i=edges[i].next){
int v=edges[i].to;
if (!pre[v]){
dfs(v);
lowlink[u]=min(lowlink[u],lowlink[v]);
} else if (!sccno[v]){
lowlink[u]=min(lowlink[u],pre[v]);
}
}
if (lowlink[u]==pre[u]){
scc_cnt++;
int x;
do{
x=stk.top();
stk.pop();
sccno[x]=scc_cnt;
}while (x!=u);
}
}
void find_scc(int n)
{
dfs_clock=scc_cnt=0;
clr(sccno,0);
clr(pre,0);
while (!stk.empty()) stk.pop();
REP_1(i,n) if (!pre[i]) dfs(i);
}
}solver;
bool a[5555][5555];
vector<int>ans;
vector<int>ot;
int main()
{
while (~scanf("%d",&n))
{
if (n==0) break;
scanf("%d",&m);
clr(a,0);
solver.init();
while (m--)
{
int x,y;
scanf("%d%d",&x,&y);
solver.addedge(x,y);
}
solver.find_scc(n);
m=solver.scc_cnt;
REP_1(u,n)
{
for (int i=solver.head[u];i!=-1;i=solver.edges[i].next)
{
int v=solver.edges[i].to;
if (solver.sccno[u]!=solver.sccno[v])
{
a[solver.sccno[u]][solver.sccno[v]]=true;
}
}
}
ans.clear();
REP_1(i,m)
{
int sum=0;
REP_1(j,m)
{
if (a[i][j]) sum++;
}
if (sum==0) ans.push_back(i);
}
ot.clear();
REP(k,sz(ans))
REP_1(i,n)
if (solver.sccno[i]==ans[k]) ot.push_back(i);
sort(ot.begin(),ot.end());
REP(i,sz(ot)-1)
{
cout<<ot[i]<<" ";
}
cout<<ot[sz(ot)-1]<<endl;
}
return 0;
}
