Poj2762 going from u to V or from V to u? --- Point reduction + Topology

To a directed graph, check whether any two points on the graph can be reached.

First, it is easy to think of shrinking to a directed acyclic graph, and second, how to deal with the reachable between any two points.

I drew some situations on the paper:

4 31 22 32 44 41 21 32 43 43 31 22 31 37 81 21 33 42 44 54 65 76 75 61 21 32 33 43 54 5NNYNY

Based on this, we have been struggling to determine how to directly determine through the inbound degree. But I always felt that it was not rigorous ..

Then we can find that the topological sequence is unique. In this way, there is no irrelevant point in the graph, and it is still very clever ..

#include<cstdio>#include<cstring>#include<cmath>#include<iostream>#include<algorithm>#include<vector>#include<queue>const int maxn=1010;const int maxm=6010;using namespace std;int h,head[maxn],n,m,in[maxn],out[maxn];int sta[maxn],top,vis[maxn],dfn[maxn],low[maxn],ccnt,id;vector<int> e[maxn];vector<int> edge[maxn];vector<int> part[maxn];int inpart[maxn];void tarjan(int x){    int i,j;    dfn[x]=low[x]=id++;    vis[x]=1;    sta[++top]=x;    for(i=0;i<e[x].size();i++)    {        j=e[x][i];        if(dfn[j]==-1)        {            tarjan(j);            low[x]=min(low[x],low[j]);        }        else if(vis[j])            low[x]=min(low[x],dfn[j]);    }    if(dfn[x]==low[x])    {        do        {            j=sta[top--];            vis[j]=0;            part[ccnt].push_back(j);            inpart[j]=ccnt;        }while(j!=x);        ccnt++;    }}void solve(){    memset(sta,-1,sizeof sta);    memset(vis,0,sizeof vis);    memset(dfn,-1,sizeof dfn);    memset(low,-1,sizeof low);    top=ccnt=id=0;    for(int i=1;i<=n;i++)            if(dfn[i]==-1)                tarjan(i);}int topo(){    queue<int> q;    for(int i=0;i<ccnt;i++)        if(in[i]==0) q.push(i);    while(!q.empty())    {        if(q.size()>1) return 0;        int x=q.front();        q.pop();        for(int i=0;i<edge[x].size();i++)        {            in[edge[x][i]]--;            if(in[edge[x][i]]==0) q.push(edge[x][i]);        }    }    return 1;}int main(){    int T,a,b,i,j;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&m);        for(i=0;i<=n;i++)        {            part[i].clear();            e[i].clear();            edge[i].clear();        }        while(m--)        {            scanf("%d%d",&a,&b);            e[a].push_back(b);        }        solve();        memset(in,0,sizeof in);        memset(out,0,sizeof out);        int flag=1;        for(i=1;i<=n;i++)        {            for(j=0;j<e[i].size();j++)            {                a=inpart[i];                b=inpart[e[i][j]];                if(a!=b)                {                    in[b]++;                    out[a]++;                    edge[a].push_back(b);                }            }        }        if(topo())            printf("Yes\n");        else printf("No\n");    }    return 0;}