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POJ-2762 Going from u to v or from v to u? 双连通分量+拓扑排序

2013-05-21 11:12 337 查看
  题目链接:http://poj.org/problem?id=2762

  判断在一个有向图中,是否任意的两点存在一条通路。

  首先用tarjan算法进行边-双连通分量缩点,接下来就是判断树的分支只有一个,那么就用拓扑排序每次判断入度为0的点是否只有一个。

//STATUS:C++_AC_360MS_340KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=1010;
const int INF=0x3f3f3f3f;
const int MOD=100000,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

struct Edge{
int u,v;
}e[N*6],e2[N*6];
int first
,next[N*6],first2
,next2[N*6],pre
,sccno
,low
,c
,vis
,p
;
int T,n,m,mt,dfs_clock,scnt,mt2;
stack<int> s;

int find(int x){return p[x]==x?x:p[x]=find(p[x]);}

void adde(int a,int b)
{
e[mt].u=a;e[mt].v=b;
next[mt]=first[a],first[a]=mt++;
}

void adde2(int a,int b)
{
e2[mt2].u=a;e2[mt2].v=b;
next2[mt2]=first2[a],first2[a]=mt2++;
}

void dfs(int u)
{
int i,j,v;
pre[u]=low[u]=++dfs_clock;
s.push(u);
for(i=first[u];i!=-1;i=next[i]){
v=e[i].v;
if(!pre[v]){
dfs(v);
low[u]=Min(low[u],low[v]);
}
else if(!sccno[v]){
low[u]=Min(low[u],low[v]);
}
}
if(low[u]==pre[u]){
int x=-1;
scnt++;
while(x!=u){
x=s.top();s.pop();
sccno[x]=scnt;
}
}
}

int topo()
{
int i,j,cnt,sum=scnt,w;
mem(c,0);
for(;sum;sum--){
mem(vis,0);
for(i=1;i<=scnt;i++){
for(j=first2[i];j!=-1;j=next2[j]){
vis[e2[j].v]=1;
}
}
cnt=0;
for(i=1;i<=scnt;i++){
if(!c[i] && !vis[i]){w=i;cnt++;}
if(cnt>=2)return 0;
}
first2[w]=-1;
c[w]=1;
}
return 1;
}

int main()
{
//   freopen("in.txt","r",stdin);
int i,j,a,b,x,y,ok;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
mem(first,-1);mt=0;
for(i=1;i<=n;i++)p[i]=i;
for(i=0;i<m;i++){
scanf("%d%d",&a,&b);
x=find(a);y=find(b);
if(x!=y)p[y]=p[x];
adde(a,b);
}
ok=0;
for(i=1;i<=n;i++){
if(p[i]==i)ok++;
if(ok>=2){ok=0;break;}
}
if(ok==1){
mem(pre,0);mem(sccno,0);
scnt=dfs_clock=0;
for(i=1;i<=n;i++){
if(!pre[i])dfs(i);
}
if(scnt!=1){
mt2=0;
mem(first2,-1);
for(i=0;i<mt;i++){
if(sccno[e[i].u]!=sccno[e[i].v]){
adde2(sccno[e[i].u],sccno[e[i].v]);
}
}
ok=topo();
}
}

printf("%s\n",ok?"Yes":"No");
}
return 0;
}
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