POJ2762 Going from u to v or from v to u? 强连通+缩点

题目链接:

poj2762

题意:

给出一幅单向图。问这张图是否满足   随意两点ab 都能 从a到达b 或  从b到达a

题解思路:

推断一幅图是否满足弱连通

首先想到的是将图中的 强连通分量(能互相到达的顶点集)  进行缩点

然后再依据原有边 又一次建图

假设缩点后的图是一条单链(回路,通路都能够)   则一定满足弱连通

推断是否是一条单链 能够依据建图过程中得到 入度 出度 数组进行推断

某点的入度 或 出度假设大于1则一定不是单链

另外单链仅仅能有一条  不能有多个点入度=0

代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#define maxn 1050
using namespace std;
struct node
{
    int to,next;
} edge[maxn*6];
int head[maxn];
int s;

int dfn[maxn],low[maxn],num;

int sta[maxn],insta[maxn],top;

int belong[maxn],block;

void init()
{
    memset(head,-1,sizeof(head));
    memset(insta,0,sizeof(insta));
    memset(dfn,0,sizeof(dfn));
    block=s=num=top=0;
}

void addedge(int a,int b)
{
    edge[s]= {b,head[a]};
    head[a]=s++;
}

void Tarjan(int u,int pre)
{
    dfn[u]=low[u]=++num;
    insta[u]=1;
    sta[top++]=u;
    for(int i=head[u]; i!=-1; i=edge[i].next)
    {
        int v=edge[i].to;
        if(!dfn[v])
        {
            Tarjan(v,u);
            low[u]=min(low[u],low[v]);
        }
        else if(insta[v])                    //回边
            low[u]=min(low[u],dfn[v]);
    }
    if(dfn[u]==low[u])                      //缩点
    {
        int d;
        block++;
        while(d!=u)
        {
            d=sta[--top];
            insta[d]=0;
            belong[d]=block;
        }
    }
}

int rebuild(int n)
{
    int indegree[maxn]= {0};
    int outdegree[maxn]={0};
    int u,v;
    for(int i=1; i<=n; i++)                            //又一次建图
    {
        u=belong[i];
        for(int j=head[i]; j!=-1; j=edge[j].next)
        {
            v=edge[j].to;
            v=belong[v];
            if(u!=v)          //不在同一个强连通分量才干建边
            {
                outdegree[u]++;
                indegree[v]++;
                if(indegree[v]>1||outdegree[u]>1)
                    return 0;
            }
        }
    }
    int ss=0;
    for(int i=1; i<=block; i++)
        if(!indegree[i])
            ss++;
    if(ss>1)             //入度=0的点有多个
        return 0;
    return 1;
}

int main()
{
    int n,m,a,b;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        init();
        scanf("%d%d",&n,&m);
        while(m--)
        {
            scanf("%d%d",&a,&b);
            addedge(a,b);
        }
        for(int i=1;i<=n;i++)   //
            if(!dfn[i])         //有向图Tarjan算法
                Tarjan(i,-1);   //

        if(rebuild_topsort(n))
            cout<<"Yes"<<endl;
        else
            cout<<"No"<<endl;
    }
    return 0;
}
时间: 2024-10-08 06:28:07

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