Cactus 【 仙人掌图 tarjan 求scc 】

Cactus
Problem Description
1. It is a Strongly Connected graph.
2. Each edge of the graph belongs to a circle and only belongs to one circle.
We call this graph as CACTUS.

There is an example as the figure above. The left one is a cactus, but the right one isn’t. Because the edge (0, 1) in the right graph belongs to two circles as (0, 1, 3) and (0, 1, 2, 3).

Input
The input consists of several test cases. The first line contains an integer T (1<=T<=10), representing the number of test cases.
For each case, the first line contains a integer n (1<=n<=20000), representing the number of points.
The following lines, each line has two numbers a and b, representing a single-way edge (a->b). Each case ends with (0 0).
Notice: The total number of edges does not exceed 50000.

Output
For each case, output a line contains “YES” or “NO”, representing whether this graph is a cactus or not.

Sample Input
2
4
0 1
1 2
2 0
2 3
3 2
0 0
4
0 1
1 2
2 3
3 0
1 3
0 0

Sample Output
YES
NO

仙人掌图形
1.没有横向边；
2.图中没有桥； // 即 图是强联通的
3.每个点的反向边和low[v]值比dfn[u]小的子节点v的数量和小于2；

用tarjan 求的时候 就是 不能够让 now点指向其下方的已经在stack中的点

代码

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<cmath>#include<queue>#include<stack>#include<map>#include<vector>#include<set>#define CLR(a,b) memset((a),(b),sizeof(a))#define inf 0x3f3f3f3f#define mod 100009#define LL long long#define M  50000+4100#define ll o<<1#define rr o<<1|1#define lson o<<1,l,mid#define rson o<<1|1,mid+1,rusing namespace std;void read(int &x){    x=0;char c;    while((c=getchar())<'0');    do x=x*10+c-'0';while((c=getchar())>='0');}struct Edge {    int from,to,next;}edge[M];int head[M],top;int low[M],dfn[M];int sccno[M],in[M];int dfs_clock,scc_cnt;vector<int>scc[M],G[M];stack<int>S;int Instack[M];int n,m;int In[M],Out[M];int flag;void addedge(int a,int b){    Edge e={a,b,head[a]};    edge[top]=e; head[a]=top++;}void init(){    memset(head,-1,sizeof(head));    top=0;    while(!S.empty()) S.pop();}void getmap(){    int a,b;    while(scanf("%d%d",&a,&b)&&(a||b))    addedge(a,b);}void tarjan(int now){    int enxts;    dfn[now]=low[now]=++dfs_clock;    S.push(now);Instack[now]=1;    for(int i=head[now];i!=-1;i=edge[i].next)    {        Edge e=edge[i];        if(!dfn[e.to])        {            tarjan(e.to);            low[now]=min(low[now],low[e.to]);        }        else if(Instack[e.to])         {            low[now]=min(low[now],dfn[e.to]);            if(low[e.to] != dfn[e.to]) flag = 1;  //  这个就是表明  当前这个点指向的点是在其下方的（并且其还在 栈内 ），只有在其下方且在栈内的 点，他的low才是已经修改过了，不等于 dfn值         }    }    if(low[now]==dfn[now])    {        scc_cnt++;        scc[scc_cnt].clear();        while(1)        {            int nexts=S.top();S.pop();Instack[nexts]=0;            sccno[nexts]=scc_cnt;            scc[scc_cnt].push_back(nexts);            if(now==nexts) break;        }    }}void find_cut(int le,int ri) {    memset(dfn,0,sizeof(dfn));    memset(low,0,sizeof(low));    memset(sccno,0,sizeof(sccno));    memset(Instack,0,sizeof(Instack));    dfs_clock=scc_cnt=0;    for(int i=0;i<n;i++)    if(!dfn[i]) tarjan(i);}void solve(){    if(scc_cnt==1&&!flag)  printf("YES\n");    else printf("NO\n");}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        flag=0;        init();        getmap();        find_cut(0,n-1);        solve();    }    return 0;}