Hdu-4635-Strongly connected-Strongly connected Component

Question:

I will give you a directed graph and ask you how many sides you can add so that the graph is still not strongly connected.

Practice:

1. Find all strongly connected components in the graph.

2. Scale the above strongly connected component into a point.

3. The question is now a complete graph. The minimum number of edges to be removed makes the graph less connected,

Therefore, it must remove all the edges of vertices with the least number of points in all strong Unicom components.

 

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>#include<stack>#include<vector>using namespace std;#define maxn 100005#define mem(a,b) memset(a,b,sizeof(a));vector<int>q[maxn];stack<int>qq;int times;int nums;int num[maxn];int dnf[maxn];int low[maxn];int out[maxn];int in[maxn];int sum[maxn];int instack[maxn];void tarjan(int x){    dnf[x]=low[x]=times++;    instack[x]=1;    qq.push(x);    int i;    for(i=0;i<q[x].size();i++)    {        int y=q[x][i];        if(!dnf[y])        {            tarjan(y);            low[x]=min(low[x],low[y]);        }        else if(instack[y])        {            low[x]=min(low[x],dnf[y]);        }    }    if(dnf[x]==low[x])    {        int y=-1;        while(y!=x)        {            y=qq.top();            qq.pop();            num[y]=nums;            instack[y]=0;            sum[nums]++;        }        nums++;    }    //cout<<x<<" "<<low[x]<<" "<<dnf[x]<<endl;}int main(){   // freopen("1004.in","r",stdin);    int T,cas,i,j,n,m;    cin>>T;    for(cas=1;cas<=T;cas++)    {        cin>>n>>m;        for(i=1;i<=n;i++)q[i].clear();        while(!qq.empty())qq.pop();        mem(in,0);        mem(out,0);        mem(dnf,0);        mem(low,0);        mem(num,0);        mem(sum,0);        mem(instack,0);        nums=0;        int x,y;        for(i=0;i<m;i++)        {            scanf("%d%d",&x,&y);            q[x].push_back(y);        }        for(i=1;i<=n;i++)            if(!dnf[i])tarjan(i);        if(nums==1)        {            printf("Case %d: -1\n",cas);            continue;        }        for(i=1;i<=n;i++)        {            for(j=0;j<q[i].size();j++)            {                if(num[i]!=num[q[i][j]])                {                    out[num[i]]++;                    in[num[q[i][j]]]++;                }            }        }        int maxx=maxn;        for(i=0;i<nums;i++)        {            if(in[i]==0||out[i]==0)            {                if(sum[i]<maxx)maxx=sum[i];            }        }        __int64 ss;        ss=0;        ss =(__int64)n*(n-1);        ss-=(__int64)m;        ss-=(__int64)maxx*(n-maxx);        printf("Case %d: %I64d\n",cas,ss);    }    return 0;}