poj-2762-Going from u to v or from v to u?-tarjan算法求缩点+算是不是一字链

来源：互联网发布：常用的网络拓扑结构是编辑：程序博客网时间：2024/06/06 02:48

tarjan求缩点，然后算缩点之后的图是不是一字链。

判断是不是一字链很简单，直接dfs求出一条最长边。

看最长边是不是等于缩点之后的数目即可。

#include <iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<stack>#include<math.h>using namespace std;#define maxm 10000#define maxn 1100#define eps 0.000001#define zero(x) ((fabs(x)<eps?0:x))#define INF 99999999struct gra{    struct list    {        int u,v,next;    } edge[maxm];    int head[maxn];    int vis[maxn];    int num;    int n;    int id[maxn];    int od[maxn];    int shuyu[maxn];    int nums;    int viss[maxn];    int dnf[maxn],low[maxn],times,instack[maxn];    stack<int>qq;    void init()    {        memset(head,-1,sizeof(head));        memset(vis,0,sizeof(vis));        memset(shuyu,0,sizeof(shuyu));        memset(dnf,0,sizeof(dnf));        memset(low,0,sizeof(low));        memset(instack,0,sizeof(instack));        memset(viss,0,sizeof(viss));        while(!qq.empty())qq.pop();        num=0;        nums=1;        times=1;    }    void add(int u,int v)    {        edge[num].u=u;        edge[num].v=v;        edge[num].next=head[u];        head[u]=num++;    }    void tarjan(int x)    {        dnf[x]=low[x]=times++;        instack[x]=1;        qq.push(x);        for(int i=head[x]; i!=-1; i=edge[i].next)        {            int y=edge[i].v;            if(!dnf[y])            {                tarjan(y);                low[x]=min(low[x],low[y]);            }            else if(instack[y])            {                low[x]=min(low[x],dnf[y]);            }        }        if(low[x]==dnf[x])        {            int y=-1;            while(x!=y)            {                y=qq.top();                shuyu[y]=nums;                qq.pop();                instack[y]=0;            }            nums++;        }    }    void start()    {        for(int i=1; i<=n; i++)        {            if(!dnf[i])tarjan(i);        }    }    int dfs(int x,int p)    {        int mm=p;        for(int i=head[x]; i!=-1; i=edge[i].next)        {            mm=max(mm,dfs(edge[i].v,p+1));        }        return mm;    }} G,g;int main(){    int n,i,j,m,a,b,T;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&m);        G.init();        g.init();        G.n=n;        for(i=1; i<=m; i++)        {            scanf("%d%d",&a,&b);            G.add(a,b);        }        G.start();        g.n=G.nums;        int nin,nout;        nin=nout=0;        for(i=1; i<=G.n; i++)        {            for(j=G.head[i]; j!=-1; j=G.edge[j].next)            {                int u=G.edge[j].u;                int v=G.edge[j].v;                if(G.shuyu[u]==G.shuyu[v])continue;                G.id[G.shuyu[v]]++;                G.od[G.shuyu[u]]++;               // cout<<G.shuyu[u]<<" - "<<G.shuyu[v]<<endl;                g.add(G.shuyu[u],G.shuyu[v]);            }        }        for(i=1; i<G.nums; i++)        {            if(G.id[i]==0)nin++;            if(G.od[i]==0)nout++;        }        if(nin>1||nout>1)        {            cout<<"No"<<endl;            continue;        }        for(i=1; i<G.nums; i++)        {            if(G.id[i]==0)break;        }      //  cout<<g.dfs(i,1)<<endl;        if(g.dfs(i,1)==G.nums-1)cout<<"Yes"<<endl;        else cout<<"No"<<endl;    }    return 0;}

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