POJ 2762 Going from u to v or from v to u?(強聯通，拓撲排序)

標籤：acm   algorithm   圖論   強聯通   拓撲排序   

http://poj.org/problem?id=2762

Going from u to v or from v to u? Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 14573   Accepted: 3849 Description In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each time, Wind choose two rooms x and y, and ask one of their little sons go from one to the other. The son can either go from x to y, or from y to x. Wind promised that her tasks are all possible, but she actually doesn‘t know how to decide if a task is possible. To make her life easier, Jiajia decided to choose a cave in which every pair of rooms is a possible task. Given a cave, can you tell Jiajia whether Wind can randomly choose two rooms without worrying about anything? Input The first line contains a single integer T, the number of test cases. And followed T cases.  The first line for each case contains two integers n, m(0 < n < 1001,m < 6000), the number of rooms and corridors in the cave. The next m lines each contains two integers u and v, indicating that there is a corridor connecting room u and room v directly.  Output The output should contain T lines. Write ‘Yes‘ if the cave has the property stated above, or ‘No‘ otherwise. Sample Input 13 31 22 33 1 Sample Output Yes Source POJ Monthly--2006.02.26,zgl & twb

題意：

給出一個有向圖，判斷對於任意兩點u,v，是否可以從u到達v或者從v到達u。

分析：

判斷有向圖的單聯通性。首先強聯通縮點，得到一個DAG，如果這個DAG是一條單鏈，那麼顯然是可以的。如何判斷DAG是否為單鏈呢？只要判斷拓撲序是否唯一即可。

/* * * Author : fcbruce <[email protected]> * * Time : Tue 14 Oct 2014 11:37:19 AM CST * */#include <cstdio>#include <iostream>#include <sstream>#include <cstdlib>#include <algorithm>#include <ctime>#include <cctype>#include <cmath>#include <string>#include <cstring>#include <stack>#include <queue>#include <list>#include <vector>#include <map>#include <set>#define sqr(x) ((x)*(x))#define LL long long#define itn int#define INF 0x3f3f3f3f#define PI 3.1415926535897932384626#define eps 1e-10#ifdef _WIN32  #define lld "%I64d"#else  #define lld "%lld"#endif#define maxm 8964#define maxn 1007using namespace std;int n,m;int fir[maxn];int u[maxm],v[maxm],nex[maxm];int e_max;int pre[maxn],low[maxn],sccno[maxn];int st[maxn],top;int scc_cnt,dfs_clock;int deg[maxn];inline void add_edge(int _u,int _v){  int e=e_max++;  u[e]=_u;v[e]=_v;  nex[e]=fir[u[e]];fir[u[e]]=e;}void tarjan_dfs(int s){  pre[s]=low[s]=++dfs_clock;  st[++top]=s;  for (int e=fir[s];~e;e=nex[e])  {    int t=v[e];    if (pre[t]==0)    {      tarjan_dfs(t);      low[s]=min(low[s],low[t]);    }    else    {      if (sccno[t]==0)        low[s]=min(low[s],pre[t]);    }  }  if (low[s]==pre[s])  {    scc_cnt++;    for (;;)    {      int x=st[top--];      sccno[x]=scc_cnt;      if (x==s) break;    }  }}void find_scc(){  scc_cnt=dfs_clock=0;  top=-1;  memset(sccno,0,sizeof sccno);  memset(pre,0,sizeof pre);  for (int i=1;i<=n;i++)    if (pre[i]==0)  tarjan_dfs(i);}bool unique_toposort(){  top=-1;  for (int i=1;i<=scc_cnt;i++)  {    if (deg[i]==0) st[++top]=i;  }  for (int i=0;i<scc_cnt;i++)  {    if (top==1 || top==-1) return false;    int x=st[top--];    for (int e=fir[x];~e;e=nex[e])    {      deg[v[e]]--;      if (deg[v[e]]==0) st[++top]=v[e];    }  }  return true;}int main(){#ifdef FCBRUCE  freopen("/home/fcbruce/code/t","r",stdin);#endif // FCBRUCE  int T_T;  scanf("%d",&T_T);  while (T_T--)  {    scanf("%d%d",&n,&m);    e_max=0;    memset(fir,-1,sizeof fir);    for (int i=0,u,v;i<m;i++)    {      scanf("%d%d",&u,&v);      add_edge(u,v);    }    find_scc();    int temp=e_max;    e_max=0;    memset(fir,-1,sizeof fir);    memset(deg,0,sizeof deg);    for (int e=0;e<temp;e++)    {      if (sccno[u[e]]==sccno[v[e]]) continue;      deg[sccno[v[e]]]++;      add_edge(sccno[u[e]],sccno[v[e]]);    }    if (unique_toposort())      puts("Yes");    else      puts("No");      }    return 0;}

POJ 2762 Going from u to v or from v to u?(強聯通，拓撲排序)